{"id":219,"date":"2026-06-11T23:35:56","date_gmt":"2026-06-11T23:35:56","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=219"},"modified":"2026-06-11T23:35:57","modified_gmt":"2026-06-11T23:35:57","slug":"work-done-in-physics","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/work-done-in-physics\/","title":{"rendered":"What Is Work Done in Physics?"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nWork done in physics is the energy transferred to or from an object when a force moves it through a distance. The formula is W = F d cos \u03b8 \u2014 force times displacement times the cosine of the angle between them. Work is a scalar quantity measured in joules (J); one joule equals one newton-metre.\n<\/p><\/div>\n\n<p>Push a broken-down car along a flat road and your aching arms are measuring something real: energy flowing out of you and into the car. Physics puts an exact number on that transfer, and the number is called work done.<\/p>\n\n<p>Now lean on the same car for ten minutes without it moving an inch. You sweat, you strain \u2014 and in physics terms you have done precisely zero work. That gap between everyday effort and scientific work is the key to the whole topic.<\/p>\n\n<h2>What Is Work Done in Physics?<\/h2>\n\n<p>Think of energy as a currency and work as the transaction that moves it between accounts. Whenever a force acts on an object while the object moves, energy changes hands \u2014 and the work done tells you exactly how many joules were transferred.<\/p>\n\n<p>The formal definition: <strong>work done is the product of the displacement and the component of the force acting along that displacement<\/strong>. A force pointing along the motion does work; a force at right angles to the motion does none, no matter how enormous it is.<\/p>\n\n<p>Work is a <strong>scalar<\/strong> \u2014 it has a size and a sign, but no direction. Its SI unit is the <strong>joule (J)<\/strong>: one joule is the work done when a one-newton force moves an object one metre in the direction of the force.<\/p>\n\n<p>The unit honours James Prescott Joule, the Manchester brewer-turned-physicist whose painstaking experiments in the 1840s showed that mechanical work and heat are two forms of the same currency: energy.<\/p>\n\n<figure style=\"margin:32px auto;max-width:420px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/05\/Joule_James_sitting.jpg\"\n       alt=\"James Prescott Joule, the physicist after whom the joule, the SI unit of work done in physics, is named\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">James Prescott Joule (1818\u20131889). His experiments showed that work and heat are both transfers of energy.<\/figcaption>\n<\/figure>\n\n<h2>The Work Done Formula: W = F d cos \u03b8<\/h2>\n\n<p>One compact equation covers every constant-force situation you will meet at GCSE, A-level and first-year university.<\/p>\n\n<div class=\"pf-formula\">W = F d cos \u03b8<\/div>\n\n<ul>\n<li><strong>W<\/strong> \u2014 work done, in joules (J)<\/li>\n<li><strong>F<\/strong> \u2014 the magnitude of the constant force, in newtons (N)<\/li>\n<li><strong>d<\/strong> \u2014 the magnitude of the displacement, in metres (m)<\/li>\n<li><strong>\u03b8<\/strong> \u2014 the angle between the force and the displacement, in degrees or radians<\/li>\n<\/ul>\n\n<p>When the force points exactly along the motion, \u03b8 = 0\u00b0 and cos \u03b8 = 1, so the equation collapses to the simpler version most students meet first:<\/p>\n\n<div class=\"pf-formula\">W = F d<\/div>\n\n<p>A quick unit check keeps everything honest: 1 J = 1 N \u00d7 1 m, and since a newton is 1 kg m\/s\u00b2, a joule is 1 kg m\u00b2\/s\u00b2. Kinetic energy, heat and electrical energy are all measured in joules too \u2014 that shared unit is the first hint that work and energy are one family.<\/p>\n\n<h3>Why the cos \u03b8?<\/h3>\n\n<p>Pull a sledge with a rope angled upwards and the rope is doing two jobs at once. Part of the tension hauls the sledge forward; part of it lifts uselessly against gravity.<\/p>\n\n<p>Only the forward part \u2014 the component F cos \u03b8 lying along the motion \u2014 actually transfers energy into the sledge&#8217;s movement. The cos \u03b8 in the formula is simply the mathematics of keeping the useful slice and discarding the rest.<\/p>\n\n<svg viewBox=\"0 0 760 430\" role=\"img\" aria-label=\"Work done in physics diagram: a crate is pulled by a force F at an angle theta above the horizontal; only the horizontal component F cos theta, which lies along the displacement d, does work\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"display:block;width:100%;height:auto;max-width:720px;margin:0 auto;\">\n  <defs>\n    <marker id=\"pfw1g\" markerUnits=\"userSpaceOnUse\" markerWidth=\"18\" markerHeight=\"18\" refX=\"13\" refY=\"9\" orient=\"auto\"><path d=\"M2 2 L16 9 L2 16 Z\" fill=\"#C8932A\"><\/path><\/marker>\n    <marker id=\"pfw1w\" markerUnits=\"userSpaceOnUse\" markerWidth=\"18\" markerHeight=\"18\" refX=\"13\" refY=\"9\" orient=\"auto\"><path d=\"M2 2 L16 9 L2 16 Z\" fill=\"#7A1F2B\"><\/path><\/marker>\n    <marker id=\"pfw1i\" markerUnits=\"userSpaceOnUse\" markerWidth=\"18\" markerHeight=\"18\" refX=\"13\" refY=\"9\" orient=\"auto\"><path d=\"M2 2 L16 9 L2 16 Z\" fill=\"#0A1628\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"0\" y=\"0\" width=\"760\" height=\"430\" rx=\"10\" fill=\"#F5F2EA\"><\/rect>\n  <text x=\"380\" y=\"44\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"21\" font-weight=\"bold\" fill=\"#0A1628\">Only the part of F along the motion does work<\/text>\n  <text x=\"380\" y=\"72\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"17\" font-style=\"italic\" fill=\"#7A1F2B\">W = F d cos \u03b8<\/text>\n  <line x1=\"60\" y1=\"330\" x2=\"700\" y2=\"330\" stroke=\"#0A1628\" stroke-width=\"3\"><\/line>\n  <rect x=\"130\" y=\"250\" width=\"120\" height=\"80\" rx=\"6\" fill=\"#142139\"><\/rect>\n  <text x=\"190\" y=\"297\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"16\" fill=\"#FAF6EE\">crate<\/text>\n  <line x1=\"250\" y1=\"270\" x2=\"422\" y2=\"270\" stroke=\"#7A1F2B\" stroke-width=\"4\" stroke-dasharray=\"9 6\" marker-end=\"url(#pfw1w)\"><\/line>\n  <line x1=\"422\" y1=\"270\" x2=\"422\" y2=\"158\" stroke=\"#142139\" stroke-width=\"2.5\" stroke-dasharray=\"6 6\"><\/line>\n  <line x1=\"250\" y1=\"270\" x2=\"422\" y2=\"150\" stroke=\"#C8932A\" stroke-width=\"5\" marker-end=\"url(#pfw1g)\"><\/line>\n  <path d=\"M 308 270 A 58 58 0 0 0 297.5 236.7\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"2\"><\/path>\n  <text x=\"316\" y=\"252\" font-family=\"Georgia, Times, serif\" font-size=\"18\" font-style=\"italic\" fill=\"#0A1628\">\u03b8<\/text>\n  <text x=\"436\" y=\"146\" font-family=\"Georgia, Times, serif\" font-size=\"22\" font-weight=\"bold\" font-style=\"italic\" fill=\"#0A1628\">F<\/text>\n  <text x=\"300\" y=\"296\" font-family=\"Georgia, Times, serif\" font-size=\"16\" font-style=\"italic\" fill=\"#7A1F2B\">F cos \u03b8 \u2014 does the work<\/text>\n  <text x=\"432\" y=\"220\" font-family=\"Georgia, Times, serif\" font-size=\"15\" font-style=\"italic\" fill=\"#142139\">F sin \u03b8 (no work)<\/text>\n  <line x1=\"150\" y1=\"370\" x2=\"560\" y2=\"370\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#pfw1i)\"><\/line>\n  <text x=\"355\" y=\"400\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"16\" font-style=\"italic\" fill=\"#0A1628\">d (displacement)<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:14px;font-style:italic;color:#1F2E47;margin-top:6px;\">Figure 1: A force at angle \u03b8 to the motion. Only the component F cos \u03b8 along the displacement transfers energy.<\/p>\n\n<h3>What if the force is not constant?<\/h3>\n\n<p>W = F d cos \u03b8 assumes the force stays steady. When it varies \u2014 a spring stretching, say \u2014 the work done becomes the <strong>area under the force\u2013displacement graph<\/strong>; at A-level that idea is written as an integral, and for a spring it gives the tidy result W = \u00bdkx\u00b2. The constant-force formula is just the special case where that area is a simple rectangle.<\/p>\n\n<h2>Positive, Negative and Zero Work<\/h2>\n\n<p>Can work done be negative? Absolutely \u2014 and the sign is doing real physics, not bookkeeping. It tells you which way the energy flowed.<\/p>\n\n<p><strong>Positive work<\/strong> means the force has a component along the motion and is feeding energy in: an engine driving a car forward, you pushing a trolley. <strong>Negative work<\/strong> means the force has a component against the motion and is draining energy out \u2014 <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">friction<\/a> under a sliding crate, brakes gripping a disc, gravity tugging on a rising ball.<\/p>\n\n<p><strong>Zero work<\/strong> happens three ways: no force, no displacement, or a force at exactly 90\u00b0 to the motion. Push a wall all afternoon (d = 0) or carry a bag at steady height across a room (support force at right angles to the motion) and the work done is nil.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Angle \u03b8<\/th>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">cos \u03b8<\/th>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Sign of work done<\/th>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Example<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">0\u00b0<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">+1<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Positive \u2014 maximum<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Horizontal push on a trolley moving forward<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">60\u00b0<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">+0.5<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Positive \u2014 reduced<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Sledge rope held at a steep angle<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">90\u00b0<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">0<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Zero<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Carrying a bag at constant height; a satellite in circular orbit<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">180\u00b0<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">\u22121<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Negative \u2014 maximum<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Friction on a sliding crate; a braking force<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<svg viewBox=\"0 0 900 330\" role=\"img\" aria-label=\"Three panels comparing positive, negative and zero work done: force in the direction of motion gives positive work, force opposite the motion gives negative work, and force at right angles to the motion gives zero work\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"display:block;width:100%;height:auto;max-width:860px;margin:0 auto;\">\n  <defs>\n    <marker id=\"pfw2i\" markerUnits=\"userSpaceOnUse\" markerWidth=\"16\" markerHeight=\"16\" refX=\"11\" refY=\"8\" orient=\"auto\"><path d=\"M2 2 L14 8 L2 14 Z\" fill=\"#0A1628\"><\/path><\/marker>\n    <marker id=\"pfw2g\" markerUnits=\"userSpaceOnUse\" markerWidth=\"16\" markerHeight=\"16\" refX=\"11\" refY=\"8\" orient=\"auto\"><path d=\"M2 2 L14 8 L2 14 Z\" fill=\"#C8932A\"><\/path><\/marker>\n    <marker id=\"pfw2w\" markerUnits=\"userSpaceOnUse\" markerWidth=\"16\" markerHeight=\"16\" refX=\"11\" refY=\"8\" orient=\"auto\"><path d=\"M2 2 L14 8 L2 14 Z\" fill=\"#7A1F2B\"><\/path><\/marker>\n    <marker id=\"pfw2b\" markerUnits=\"userSpaceOnUse\" markerWidth=\"16\" markerHeight=\"16\" refX=\"11\" refY=\"8\" orient=\"auto\"><path d=\"M2 2 L14 8 L2 14 Z\" fill=\"#142139\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"0\" y=\"0\" width=\"900\" height=\"330\" rx=\"10\" fill=\"#F5F2EA\"><\/rect>\n  <text x=\"450\" y=\"40\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"19\" font-weight=\"bold\" fill=\"#0A1628\">The angle between force and motion decides the sign<\/text>\n  <rect x=\"15\" y=\"62\" width=\"270\" height=\"248\" rx=\"8\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"><\/rect>\n  <text x=\"150\" y=\"95\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"17\" font-weight=\"bold\" fill=\"#0A1628\">Positive work<\/text>\n  <text x=\"150\" y=\"138\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"30\" font-weight=\"bold\" fill=\"#C8932A\">W &gt; 0<\/text>\n  <text x=\"150\" y=\"173\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">motion<\/text>\n  <line x1=\"55\" y1=\"185\" x2=\"225\" y2=\"185\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#pfw2i)\"><\/line>\n  <line x1=\"55\" y1=\"225\" x2=\"225\" y2=\"225\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#pfw2g)\"><\/line>\n  <text x=\"150\" y=\"250\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">force<\/text>\n  <text x=\"150\" y=\"287\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" font-style=\"italic\" fill=\"#0A1628\">e.g. pushing a trolley forward<\/text>\n  <rect x=\"315\" y=\"62\" width=\"270\" height=\"248\" rx=\"8\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"><\/rect>\n  <text x=\"450\" y=\"95\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"17\" font-weight=\"bold\" fill=\"#0A1628\">Negative work<\/text>\n  <text x=\"450\" y=\"138\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"30\" font-weight=\"bold\" fill=\"#7A1F2B\">W &lt; 0<\/text>\n  <text x=\"450\" y=\"173\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">motion<\/text>\n  <line x1=\"355\" y1=\"185\" x2=\"525\" y2=\"185\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#pfw2i)\"><\/line>\n  <line x1=\"525\" y1=\"225\" x2=\"355\" y2=\"225\" stroke=\"#7A1F2B\" stroke-width=\"4\" marker-end=\"url(#pfw2w)\"><\/line>\n  <text x=\"450\" y=\"250\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">force<\/text>\n  <text x=\"450\" y=\"287\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" font-style=\"italic\" fill=\"#0A1628\">e.g. friction on a sliding crate<\/text>\n  <rect x=\"615\" y=\"62\" width=\"270\" height=\"248\" rx=\"8\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"><\/rect>\n  <text x=\"750\" y=\"95\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"17\" font-weight=\"bold\" fill=\"#0A1628\">Zero work<\/text>\n  <text x=\"750\" y=\"138\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"30\" font-weight=\"bold\" fill=\"#142139\">W = 0<\/text>\n  <text x=\"750\" y=\"173\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">motion<\/text>\n  <line x1=\"655\" y1=\"185\" x2=\"825\" y2=\"185\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#pfw2i)\"><\/line>\n  <line x1=\"740\" y1=\"262\" x2=\"740\" y2=\"208\" stroke=\"#142139\" stroke-width=\"4\" marker-end=\"url(#pfw2b)\"><\/line>\n  <text x=\"756\" y=\"240\" font-family=\"Georgia, Times, serif\" font-size=\"13\" fill=\"#0A1628\">force<\/text>\n  <text x=\"750\" y=\"287\" text-anchor=\"middle\" font-family=\"Georgia, Times, serif\" font-size=\"13\" font-style=\"italic\" fill=\"#0A1628\">e.g. carrying a bag at steady height<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:14px;font-style:italic;color:#1F2E47;margin-top:6px;\">Figure 2: Positive, negative and zero work \u2014 same formula, three very different stories.<\/p>\n\n<p>Want to feel the ang\u00adle effect rather than read about it? The lab below lets you set the force, the angle and the distance, then watch the work done respond \u2014 including the moment it hits zero at 90\u00b0 and turns negative beyond it.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Work Done Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame.pf-work{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame.pf-work{height:1000px}}<\/style><iframe src=\"\/labs\/work-done.html\" class=\"pf-sim-frame pf-work\" loading=\"lazy\"><\/iframe><\/div><\/div>\n<h2>How to Calculate Work Done in 4 Steps<\/h2>\n<p>Every constant-force problem yields to the same routine. Train the habit now and the harder questions later become mere bookkeeping.<\/p>\n<ol>\n<li><strong>Identify the force.<\/strong> Write down its magnitude in newtons and the direction it acts. If several forces act, decide whose work the question wants \u2014 or whether it wants the net work.<\/li>\n<li><strong>Identify the displacement.<\/strong> Magnitude in metres, plus direction. No displacement means no work \u2014 you can stop right there.<\/li>\n<li><strong>Find the angle \u03b8 between them.<\/strong> Between the force and the displacement \u2014 not between the force and the vertical, which is a classic slip.<\/li>\n<li><strong>Multiply and sign it.<\/strong> W = F d cos \u03b8. Attach joules, and check the sign matches the physics: energy fed in should be positive, energy drained should be negative.<\/li>\n<\/ol>\n<p>A common student slip is multiplying the full force by the distance when only a component acts along the motion. The rope-at-an-angle questions in the worked section below exist to break that habit.<\/p>\n<p>Sanity-check your magnitudes, too. Lifting a one-litre bottle of water through one metre takes about 10 J, and a 70 kg person climbing one flight of stairs does roughly 2,000 J against gravity. If your answer for a pushed shopping trolley comes out at two megajoules, something has slipped.<\/p>\n<h2>Real-World Examples of Work Done<\/h2>\n<h3>1. Pushing a supermarket trolley<\/h3>\n<p>A steady 20 N push along a 30 m aisle transfers W = 20 \u00d7 30 = 600 J into the trolley \u2014 most of it promptly stolen back by friction. Positive work in, quietly dissipating as gentle heat in the wheels and floor.<\/p>\n<h3>2. Lifting weights at the gym<\/h3>\n<p>Raising a 60 kg barbell through half a metre is W = mgh = 60 \u00d7 9.81 \u00d7 0.5 \u2248 294 J per lift. Lower it under control and gravity does +294 J on the bar while your muscles do negative work \u2014 which is why the lowering phase still burns.<\/p>\n<h3>3. A car braking<\/h3>\n<p>A 1,200 kg car at 13 m\/s (about 47 km\/h) carries roughly 100,000 J of kinetic energy. To stop it, friction at the brakes must do \u2212100,000 J of work, turning all that motion into heat in the discs. Negative work is not a technicality \u2014 it is how every vehicle stops.<\/p>\n<h3>4. A crane lifting a steel beam<\/h3>\n<p>Hoisting a 2,000 kg beam 30 m to the top of a building takes W = 2,000 \u00d7 9.81 \u00d7 30 \u2248 590,000 J \u2014 about 590 kJ banked as gravitational potential energy. That energy has not vanished; release the beam and every joule comes back as kinetic energy.<\/p>\n<h3>5. A satellite in circular orbit<\/h3>\n<p>Earth&#8217;s gravity pulls on a satellite constantly, yet in a circular orbit the pull is always at 90\u00b0 to the motion. Cos 90\u00b0 = 0, so gravity does no work at all \u2014 which is exactly why the satellite&#8217;s speed never changes. A huge force, acting forever, transferring nothing.<\/p>\n<h2>The Work\u2013Energy Theorem: What Work Actually Does<\/h2>\n<p>So far we have calculated work; the work\u2013energy theorem tells you what it buys. The result is one of the most useful shortcuts in mechanics.<\/p>\n<div class=\"pf-formula\">W_net = \u0394KE = \u00bdmv\u00b2 \u2212 \u00bdmu\u00b2<\/div>\n<ul>\n<li><strong>W_net<\/strong> \u2014 the net (total) work done on the object by all forces, in joules (J)<\/li>\n<li><strong>m<\/strong> \u2014 mass, in kilograms (kg)<\/li>\n<li><strong>u<\/strong> \u2014 initial speed, in metres per second (m\/s)<\/li>\n<li><strong>v<\/strong> \u2014 final speed, in metres per second (m\/s)<\/li>\n<\/ul>\n<p>In words: the net work done on an object equals its change in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a>. Positive net work speeds it up, negative net work slows it down, and zero net work leaves the speed untouched \u2014 however violent the individual forces are.<\/p>\n<h3>Where the theorem comes from<\/h3>\n<p>For a constant net force along the motion, <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a> gives a = F\/m, and the kinematic equation v\u00b2 = u\u00b2 + 2ad supplies the link to distance.<\/p>\n<p>Rearrange the kinematics to a d = (v\u00b2 \u2212 u\u00b2)\/2, then multiply Newton&#8217;s law by d: F d = m a d = \u00bdmv\u00b2 \u2212 \u00bdmu\u00b2. The left side is the net work; the right side is the change in kinetic energy. Two views of motion \u2014 forces and energy \u2014 joined by one line of algebra.<\/p>\n<h3>Work done by gravity and stored energy<\/h3>\n<p>Near Earth&#8217;s surface, gravity&#8217;s work depends only on the vertical height change: W = mgh, positive on the way down, negative on the way up, whatever route is taken. Forces with that path-independent property are called conservative.<\/p>\n<p>Lift an object and the work you do against gravity is banked as gravitational potential energy, E_p = mgh. Let it fall and gravity pays the balance back as kinetic energy \u2014 and you can watch the exchange happen live below.<\/p>\n<p>Set the drop height and follow the readouts: every joule of potential energy the object loses is a joule of work gravity has done on it, reappearing as kinetic energy. The total never wavers.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Energy Conservation Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame.pf-energy{width:100%;border:none;height:560px}@media(max-width:760px){.pf-sim-frame.pf-energy{height:840px}}<\/style><iframe src=\"\/labs\/energy.html\" class=\"pf-sim-frame pf-energy\" loading=\"lazy\"><\/iframe><\/div><\/div>\n<h2>Work Done vs Power: What Is the Difference?<\/h2>\n<p>Work measures how much energy moved; power measures how fast it moved. Carry bricks upstairs in one trip or ten and the work against gravity is identical \u2014 the single trip simply demands far more power.<\/p>\n<div class=\"pf-formula\">P = W \/ t<\/div>\n<p>Here P is power in watts (W), W is the work done in joules and t is the time taken in seconds. One watt is one joule per second. And yes, physics reused the letter: an italic W in equations means work, while W after a number means watts.<\/p>\n<p>For a 70 kg person climbing a 3 m flight of stairs, the work against gravity is about 2,060 J. Sprint it in 4 s and your useful power output is roughly 515 W; stroll it in 12 s and it drops to about 172 W. Same work, different power.<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Aspect<\/th>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Work done<\/th>\n<th style=\"background:#0A1628;color:#FAF6EE;padding:10px 12px;text-align:left;\">Power<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>Definition<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Energy transferred by a force acting through a distance<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Rate of transferring energy (work per unit time)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>Formula<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">W = F d cos \u03b8<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">P = W \/ t (or P = F v)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>SI unit<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">joule (J)<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">watt (W), where 1 W = 1 J\/s<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>Type<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Scalar (can be negative)<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">Scalar<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>Asks<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">How much energy moved?<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">How quickly did it move?<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\"><strong>Example<\/strong><\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">2,060 J climbing a staircase<\/td>\n<td style=\"padding:10px 12px;border-bottom:1px solid #D9CFB8;\">515 W climbing it in 4 s<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<h2>Common Misconceptions About Work Done<\/h2>\n<h3>&#8220;If I push hard, I am doing work&#8221;<\/h3>\n<p>Not unless something moves. Shove an immovable wall and the displacement is zero, so W = F d cos \u03b8 = 0, however hard you push. Your muscles burn chemical energy keeping their fibres firing \u2014 but none of it is transferred to the wall as mechanical work.<\/p>\n<h3>&#8220;Carrying a bag across a room does work on the bag&#8221;<\/h3>\n<p>Walk at steady speed and constant height, and the force you exert on the bag points straight up while the bag moves horizontally. That is a 90\u00b0 angle, so your supporting force does zero work. Examiners adore this question precisely because intuition screams otherwise.<\/p>\n<p>(Strictly, the small accelerations as you start, stop and bob do tiny bits of work \u2014 but for the steady walk, the textbook answer is zero.)<\/p>\n<h3>&#8220;A bigger force always means more work&#8221;<\/h3>\n<p>Only the component along the motion counts. A 1,000 N force at 90\u00b0 to the displacement does nothing, while a 50 N force along it does plenty. Distance matters as much as strength: half the force over double the distance transfers exactly the same energy.<\/p>\n<h3>&#8220;Negative work means a mistake&#8221;<\/h3>\n<p>Negative work is physics working perfectly: it means energy is being removed from the object. Brakes, friction, air resistance and gravity-on-the-way-up all do negative work. Without it, nothing could ever slow down.<\/p>\n<h2>How Work Done Connects to the Rest of Physics<\/h2>\n<p>Work is the hinge between two great descriptions of motion. On one side sit forces and Newton&#8217;s laws, predicting acceleration instant by instant; on the other sits <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-energy-in-physics\/\">energy<\/a>, tracking what the universe owes and is owed. Multiply a force by the distance it acts over and you cross from one picture into the other.<\/p>\n<p>Follow the energy trail further and you reach thermodynamics. The negative work friction does on a sliding crate does not destroy energy \u2014 it converts ordered motion into the disordered jiggling we measure as <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/heat-vs-temperature\/\">heat<\/a>. Energy is conserved; usefulness is not.<\/p>\n<p>One contrast is worth memorising. Force \u00d7 time is impulse, which changes momentum; force \u00d7 distance is work done, which changes kinetic energy. Two different multiplications of the same force, answering two different questions.<\/p>\n<p>If you enjoy exploring by concept map, Georgia State University&#8217;s <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/HPhc.html\" target=\"_blank\" rel=\"noopener\">HyperPhysics<\/a> project links work outward to the whole of mechanics \u2014 a fine second telling of everything on this page.<\/p>\n<h2>Worked Problems<\/h2>\n<p>Work through these in order \u2014 each adds one new idea. Cover the solutions and attempt every problem cold first.<\/p>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A warehouse worker pushes a crate 4.0 m across a level floor with a steady horizontal force of 50 N. How much work does the worker do on the crate?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: The force and the motion point the same way, so \u03b8 = 0\u00b0 and cos \u03b8 = 1. Use W = F d.\nStep 2: Substitute with units: W = 50 N \u00d7 4.0 m.\nStep 3: Solve: W = 200 J.\n<strong>Answer: W = 200 J (2.0 \u00d7 10\u00b2 J to 2 significant figures)<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A child pulls a sledge 25 m across flat snow using a rope held at 30\u00b0 above the horizontal. The tension in the rope is 80 N. Calculate the work done by the child on the sledge.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: The force is at an angle to the motion, so W = F d cos \u03b8.\nStep 2: Substitute with units: W = 80 N \u00d7 25 m \u00d7 cos 30\u00b0 = 2,000 J \u00d7 0.866.\nStep 3: Solve: W = 1,732 J.\n<strong>Answer: W \u2248 1.7 \u00d7 10\u00b3 J (about 1.7 kJ)<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A waiter carries a 2.0 kg tray 12 m across a restaurant at constant speed and constant height. How much work does the waiter&#039;s upward supporting force do on the tray?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: The supporting force is vertical (about mg \u2248 19.6 N upward); the displacement is horizontal.\nStep 2: The angle between them is 90\u00b0, and cos 90\u00b0 = 0.\nStep 3: W = 19.6 N \u00d7 12 m \u00d7 0 = 0.\n<strong>Answer: W = 0 J \u2014 a force at right angles to the motion does no work; the mass and distance are red herrings<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A removal worker lifts a 12 kg box from the floor onto a shelf 1.5 m above, raising it at constant speed. Take g = 9.81 m\/s\u00b2. How much work does the worker do against gravity?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Constant speed means the lifting force just balances the weight: F = mg = 12 kg \u00d7 9.81 m\/s\u00b2 = 117.7 N.\nStep 2: Force and motion are both vertically upward, so \u03b8 = 0\u00b0: W = F d = 117.7 N \u00d7 1.5 m.\nStep 3: Solve: W = 176.6 J.\n<strong>Answer: W \u2248 1.8 \u00d7 10\u00b2 J (about 177 J), now stored as gravitational potential energy<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A 6.0 kg crate slides 8.0 m across a horizontal floor where the coefficient of kinetic friction is 0.25. Take g = 9.81 m\/s\u00b2. Calculate the work done by friction on the crate.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: On a level floor the normal force is N = mg = 6.0 kg \u00d7 9.81 m\/s\u00b2 = 58.9 N.\nStep 2: Friction force: f = \u03bcN = 0.25 \u00d7 58.9 N = 14.7 N, acting opposite the motion, so \u03b8 = 180\u00b0 and cos \u03b8 = \u22121.\nStep 3: W = 14.7 N \u00d7 8.0 m \u00d7 (\u22121) = \u2212117.7 J.\n<strong>Answer: W \u2248 \u22121.2 \u00d7 10\u00b2 J (about \u2212118 J) \u2014 the negative sign shows friction removed energy from the crate, which became heat<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A 1,200 kg car accelerates from 10 m\/s to 25 m\/s along a straight road. Use the work\u2013energy theorem to find the net work done on the car.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: The theorem: W_net = \u00bdm(v\u00b2 \u2212 u\u00b2).\nStep 2: Substitute with units: W_net = \u00bd \u00d7 1,200 kg \u00d7 ((25 m\/s)\u00b2 \u2212 (10 m\/s)\u00b2) = 600 kg \u00d7 (625 \u2212 100) m\u00b2\/s\u00b2.\nStep 3: Solve: W_net = 600 \u00d7 525 = 315,000 J.\n<strong>Answer: W_net \u2248 3.2 \u00d7 10\u2075 J (about 315 kJ)<\/strong>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 15 kg crate, initially at rest, is dragged 10 m across a horizontal floor by a 90 N force directed 25\u00b0 above the horizontal. The coefficient of kinetic friction is 0.20 and g = 9.81 m\/s\u00b2. Find (a) the net work done on the crate and (b) its final speed.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Work by the applied force: W_F = F d cos \u03b8 = 90 N \u00d7 10 m \u00d7 cos 25\u00b0 = 900 \u00d7 0.906 = 815.7 J.\nStep 2: Normal force (the upward pull lightens the crate): N = mg \u2212 F sin 25\u00b0 = 147.15 N \u2212 38.0 N = 109.1 N.\nStep 3: Friction: f = \u03bcN = 0.20 \u00d7 109.1 N = 21.8 N opposing motion, so W_f = \u221221.8 N \u00d7 10 m = \u2212218.2 J.\nStep 4: Gravity and the normal force act at 90\u00b0 to the motion and do zero work, so W_net = 815.7 \u2212 218.2 = 597.5 J.\nStep 5: Work\u2013energy theorem with u = 0: \u00bdmv\u00b2 = W_net, so v = \u221a(2 \u00d7 597.5 J \/ 15 kg) = \u221a79.7 = 8.9 m\/s.\n<strong>Answer: (a) W_net \u2248 6.0 \u00d7 10\u00b2 J (about 598 J); (b) v \u2248 8.9 m\/s<\/strong>\n<\/div><\/details><\/div>\n<h2>Frequently Asked Questions<\/h2>\n<details class=\"pf-faq-item\"><summary>What is work done in physics?<\/summary><div class=\"pf-faq-item-answer\">\nWork done is the energy transferred when a force moves an object through a distance. It equals the component of the force along the motion multiplied by the distance moved: W = F d cos \u03b8. Work is a scalar measured in joules; one joule is the work done by a one-newton force acting through one metre.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the SI unit of work done?<\/summary><div class=\"pf-faq-item-answer\">\nThe SI unit of work done is the joule (J). One joule equals one newton-metre \u2014 the work done when a force of one newton moves an object one metre in the force&#8217;s direction. In base units, 1 J = 1 kg m\u00b2\/s\u00b2. Larger amounts are usually quoted in kilojoules (kJ) or megajoules (MJ).\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Can work done be negative?<\/summary><div class=\"pf-faq-item-answer\">\nYes. Work done is negative whenever the force, or its component, points opposite to the displacement \u2014 the angle \u03b8 exceeds 90\u00b0, making cos \u03b8 negative. Friction on a sliding box and gravity on a rising ball both do negative work. Negative work removes kinetic energy from an object, which is exactly how brakes stop a car.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Why is no work done when you hold a heavy bag still?<\/summary><div class=\"pf-faq-item-answer\">\nBecause nothing moves: the displacement is zero, and W = F d cos \u03b8 gives zero whenever d = 0. Your muscles still consume chemical energy \u2014 their fibres repeatedly contract, which is why you tire \u2014 but none of that energy is transferred to the bag as mechanical work. Physics work needs both a force and movement along it.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Is work done a scalar or a vector quantity?<\/summary><div class=\"pf-faq-item-answer\">\nWork done is a scalar. Although it is built from two vectors \u2014 force and displacement \u2014 their scalar (dot) product produces a single number with a sign but no direction. The sign carries meaning: positive work adds energy to an object, negative work removes it. So work can be +200 J or \u2212200 J, but never 200 J north.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the difference between work done and power?<\/summary><div class=\"pf-faq-item-answer\">\nWork done measures how much energy is transferred; power measures how fast it is transferred. Power equals work divided by time, P = W\/t, and is measured in watts, where 1 W = 1 J\/s. Climbing the same staircase quickly or slowly involves the same work against gravity, but the faster climb develops more power.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>How is work done related to energy?<\/summary><div class=\"pf-faq-item-answer\">\nWork is the transfer mechanism for mechanical energy. When net work is done on an object, its kinetic energy changes by exactly that amount \u2014 the work\u2013energy theorem, W_net = \u0394KE. Lifting an object stores the work done against gravity as gravitational potential energy. In short: energy is the quantity, work is the transaction.\n<\/div><\/details>\n<p>Work done is the quiet bookkeeping behind every push, lift, brake and orbit: a force, a distance, an angle, and energy on the move. Master the sign convention and the component idea, and the rest of energy physics opens up from here.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Work done in physics is the energy a force transfers when it moves an object. This guide explains the W = F d cos \u03b8 formula with real examples, seven worked problems and an interactive lab.<\/p>\n","protected":false},"author":1,"featured_media":220,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[89,21,45,88,47],"class_list":["post-219","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics","tag-energy-transfer","tag-force","tag-mechanics","tag-work-done","tag-work-energy-theorem"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/219","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=219"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/219\/revisions"}],"predecessor-version":[{"id":221,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/219\/revisions\/221"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/220"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=219"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=219"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=219"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}