{"id":1989,"date":"2020-12-18T09:46:09","date_gmt":"2020-12-18T04:16:09","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=1989"},"modified":"2020-12-18T11:31:36","modified_gmt":"2020-12-18T06:01:36","slug":"work-in-physics","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/work-in-physics\/","title":{"rendered":"How do you find work in physics?"},"content":{"rendered":"<h2>What is the formula for work?<\/h2>\n<p><strong>Definition:<\/strong> In our daily life &#8220;work&#8221; implies an activity resulting in muscular or mental exertion. However, in physics the term &#8216;work&#8217; is used in a specific sense involves the <a href=\"https:\/\/cbselibrary.com\/difference-between-distance-and-displacement\/\">displacement<\/a> of a particle or body under the action of a <a href=\"https:\/\/cbselibrary.com\/force\/\">force<\/a>. &#8220;work is said to be done when the point of application of a force moves.<br \/>\nWork done in moving a body is equal to the product of force exerted on the body and the <a href=\"https:\/\/cbselibrary.com\/difference-between-distance-and-displacement\/\">distance<\/a> moved by the body in the direction of force.<br \/>\n<strong>Work = Force \u00d7 Distance moved in the direction of force<\/strong>.<\/p>\n<p><strong>The work done by a force on a body depends on two factors<\/strong><br \/>\n(i) Magnitude of the force, and<br \/>\n(ii) Distance through which the body moves (in the direction of force)<\/p>\n<p><strong>Unit of Work<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-1990 size-full\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-and-Energy.png\" alt=\"Work-and-Energy\" width=\"337\" height=\"176\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-and-Energy.png 337w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-and-Energy-300x157.png 300w\" sizes=\"(max-width: 337px) 100vw, 337px\" \/><br \/>\nWhen a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule.<br \/>\nWork = Force \u00d7 Displacement<br \/>\n1 joule = 1 N \u00d7 1 m<br \/>\nor 1 J = 1 Nm (In SI unit)<\/p>\n<h3><strong>Work Done Analysis <\/strong><\/h3>\n<p><strong>Work done when force and displacement are along same line:<\/strong><br \/>\n<strong>Work done by a force:<\/strong> Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.<br \/>\n<strong>Work done against the force:<\/strong> Work is said to be done against a force if the direction of the displacement is opposite to that of the force.<br \/>\n<strong>Work done against gravity:<\/strong> To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If &#8216;m&#8217; is the mass of the object and &#8216;h&#8217; is the height through which it is raised, then the upward force<br \/>\n(F) = force of gravity = mg<br \/>\nIf &#8216;W&#8217; stands for work done, then<br \/>\nW = F . h = mg . h<br \/>\nThus W = mgh<br \/>\nTherefore we can say that, &#8220;The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted.<\/p>\n<p><strong> Work done when force and displacement are inclined (Oblique case):<\/strong><br \/>\nConsider a force &#8216;F&#8217; acting at angle \u03b8 to the direction of displacement &#8216;s&#8217; as shown in fig.<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-1996 size-full\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-done.png\" alt=\"Work-done\" width=\"411\" height=\"154\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-done.png 411w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Work-done-300x112.png 300w\" sizes=\"(max-width: 411px) 100vw, 411px\" \/><br \/>\nWork done when force is perpendicular to Displacement<br \/>\n\u03b8 = 90\u00ba<br \/>\nW = F.S \u00d7 cos 90\u00ba = F.S \u00d7 0 = 0<br \/>\nThus no work is done when a force acts at right angle to the displacement.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Special Examples<\/strong><\/p>\n<ol>\n<li>When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.<\/li>\n<li>Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.<\/li>\n<li>A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.<\/li>\n<\/ol>\n<h2><strong>Work Done By A Force Example Problems With Solutions<\/strong><\/h2>\n<p><strong>Example 1: \u00a0\u00a0<\/strong> How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0The work done is calculated by using the formula:<br \/>\nW = F \u00d7 S<br \/>\nHere, \u00a0 \u00a0 \u00a0Force F = 10 N<br \/>\nAnd, Distance, S = 1 m<br \/>\nSo, Work done, W = 10 \u00d7 1 J<br \/>\n= 10 J<br \/>\nThus, the work done is 10 joules<\/p>\n<p><strong>Example 2: \u00a0\u00a0<\/strong> Find the work done by a force of 10 N in moving an object through a distance of 2 m.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0 Work done = Force \u00d7 Distance moved<br \/>\nHere, \u00a0 \u00a0 \u00a0 \u00a0 Force = 10 N<br \/>\nDistance moved = 2 m<br \/>\nWork done, W = 10 N \u00d7 2 m<br \/>\n= 20 Joule = 20 J<\/p>\n<p><strong>Example 3: \u00a0\u00a0<\/strong> Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Force, F = 120 N; Distance, s = 100 m<br \/>\nUsing the formula, we have<br \/>\nW = Fs = 120 N \u00d7 100 m = 12,000 J<\/p>\n<p><strong>Example 4: \u00a0\u00a0<\/strong> A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m\/s<sup>2<\/sup>. Calculate the work done.<br \/>\n<strong>Sol.<\/strong> Given: mass, m = 5 kg<br \/>\nacceleration, a = 3 m\/s<sup>2<\/sup><br \/>\nForce acting on the body is given by<br \/>\nF = ma = 5 \u00d7 3 = 15 N<br \/>\nNow, work done is given by<br \/>\nW = Fs = 15 N \u00d7 4 m = 60 J<\/p>\n<p><strong>Example 5: \u00a0\u00a0<\/strong> Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms<sup>-2<\/sup>).<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Force of gravity<br \/>\nmg = 200 \u00d7 9.8 = 1960.0 N<br \/>\nh = 5 m<br \/>\nWork done, W = mgh<br \/>\nor W = 1960 \u00d7 5 = 9800 J<\/p>\n<p><strong>Example 6: \u00a0 \u00a0<\/strong>A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60\u00ba with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Given F = 100 N, s = 3 m, \u03b8= 60\u00ba.<br \/>\nWork done is given by<br \/>\nW = Fs cos \u03b8= 100 \u00d7 2 \u00d7 cos 60\u00ba<br \/>\n= 100 \u00d7 3 \u00d7 1\/2 = 150 J (\u2235 cos 60\u00ba = 1\/2 )<\/p>\n<p><strong>Example 7: \u00a0\u00a0<\/strong>\u00a0An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Given W = 64,000 J; F = 8,000 N<br \/>\nWork done is given by W = Fs<br \/>\nor \u00a0 \u00a064000 = 8000 \u00d7 s<br \/>\nor \u00a0 \u00a0s = 8 m<\/p>\n<p><strong>Example 8: \u00a0\u00a0<\/strong>\u00a0A horse applying a force of 800 N in pulling a cart with a constant speed of 20 m\/s. Calculate the power at which horse is working.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Power, P is given by force \u00d7 velocity, i.e.<br \/>\nP = F . v<br \/>\nHere F = 800 N; v = 20 m\/s<br \/>\n\u2234 P = 800 \u00d7 20 = 16000 watt<br \/>\n= 16 kW<\/p>\n<p><strong>Example 9: \u00a0\u00a0<\/strong>\u00a0A boy keeps on his palm a mass of 0.5 kg. He lifts the palm vertically by a distance of 0.5 m. Calculate the amount of work done.<br \/>\nUse g = 9.8 m\/s<sup>2<\/sup>.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Work done, W = F . s<br \/>\nHere, force F of gravity applied to lift the mass, is given by<br \/>\nF = mg<br \/>\n= (0.5 kg) \u00d7 (9.8 m\/s<sup>2<\/sup>)<br \/>\n= 4.9 N<br \/>\nand s = 0.5 m<br \/>\nTherefore, W = (4.9) . (0.5m) = 2.45 J.<\/p>\n<p><strong>Example 10: \u00a0\u00a0<\/strong>\u00a0A truck of mass 2500 kg is stopped by a force of 1000 N. It stops at a distance of 320 m. What is the amount of work done ? Is the work done by the force or against the force?<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0Here the force, F = 1000 N<br \/>\nDisplacement, s = 320 m<br \/>\n\u2234 Work done, W = F . s<br \/>\n= (1000N) . (320 m)<br \/>\n= 320000 J<br \/>\nIn this case, the force acts opposite to the direction of displacement. So the work is done against the force.<\/p>\n<p><strong>Example 11: \u00a0\u00a0<\/strong>\u00a0A car weighing 1200 kg and travelling at a speed of 20 m\/s stops at a distance of 40 m retarding uniformly. Calculate the force exerted by the brakes. Also calculate the work done by the brakes.<br \/>\n<strong>Solution: \u00a0\u00a0<\/strong>\u00a0In order to calculate the force applied by the brakes, we first calculate the retardation.<br \/>\nInitial speed, u = 20 m\/s; final speed,<br \/>\nv = 0, distance covered, s = 90 m<br \/>\nUsing the equation, v<sup>2<\/sup> = u<sup>2<\/sup> + 2as, we get<br \/>\n0<sup>2<\/sup> = (20)<sup>2<\/sup> + 2 \u00d7 a \u00d7 40<br \/>\nor \u00a0 80a = \u2013400<br \/>\nor \u00a0 a = \u20135 m\/s<sup>2<\/sup><br \/>\nForce exerted by the brakes is given by<br \/>\nF = ma<br \/>\nHere m = 1200 kg; a = \u2013 5 m\/s<sup>2<\/sup><br \/>\n\u2234 F = 1200 \u00d7 (\u20135) = \u2013 6000 N<br \/>\nThe negative sign shows that it is a retarding force.<br \/>\nNow, the work done by the brakes is given by<br \/>\nW = Fs<br \/>\nHere F = 6000 N; s = 40 m<br \/>\n\u2234 W = 6000 \u00d7 40 J = 240000 J<br \/>\n= 2.4 \u00d7 10<sup>5<\/sup> J<br \/>\n\u2234 Work done by the brakes = 2.4 \u00d7 10<sup>5<\/sup> J<\/p>\n<p><strong>Example 12.\u00a0<\/strong>A worker uses a horizontal force of 400 N to push a rubbish cart.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-89648\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-1.png\" alt=\"work 1\" width=\"257\" height=\"113\" \/><br \/>\nCalculate the work done to push the cart a distance of 180 m on a horizontal surface.<br \/>\n<strong>Solution:<\/strong><br \/>\nSince the direction of the movement of the cart is the same as the direction of the applied force, therefore W = F x s<br \/>\n= 400 x 180 = 72 000 J<\/p>\n<p><strong>Example 13.<\/strong> A boy uses a force of 120 N to pull a crate along a straight corridor. The applied force is at an angle of 30\u00b0 with the horizontal floor.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-89651\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-2.png\" alt=\"work 2\" width=\"312\" height=\"119\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-2.png 312w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-2-300x114.png 300w\" sizes=\"(max-width: 312px) 100vw, 312px\" \/><br \/>\nCalculate the work done by him after pulling the crate a distance of 90 m.<br \/>\n<strong>Solution:<\/strong><br \/>\nW = F cos \u03b8\u00a0x s<br \/>\n= (120 cos 30\u00b0) x 90 = 9353 J<\/p>\n<p><strong>Example 14.<\/strong> Jin uses a force of 16 N to mop a floor. The applied force makes an angle of 28\u00b0 with the floor.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-89654\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-3.png\" alt=\"work 3\" width=\"302\" height=\"147\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-3.png 302w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/work-3-300x146.png 300w\" sizes=\"(max-width: 302px) 100vw, 302px\" \/><br \/>\nWhat is the work done by Jin when he pushes the mop a horizontal distance of 2.5 m?<br \/>\n<strong>Solution:<\/strong><br \/>\nW = F cos \u03b8\u00a0x s<br \/>\n= (16 cos 28\u00b0) x 2.5 = 35.3 J<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the formula for work? Definition: In our daily life &#8220;work&#8221; implies an activity resulting in muscular or mental exertion. However, in physics the term &#8216;work&#8217; is used in a specific sense involves the displacement of a particle or body under the action of a force. &#8220;work is said to be done when the &#8230; <a title=\"How do you find work in physics?\" class=\"read-more\" href=\"https:\/\/cbselibrary.com\/work-in-physics\/\" aria-label=\"More on How do you find work in physics?\">Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[404],"tags":[759,760,758,771,772],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v18.4.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>How do you find work in physics? - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/work-in-physics\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How do you find work in physics?\" \/>\n<meta property=\"og:description\" content=\"What is the formula for work? 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Definition: In our daily life &#8220;work&#8221; implies an activity resulting in muscular or mental exertion. However, in physics the term &#8216;work&#8217; is used in a specific sense involves the displacement of a particle or body under the action of a force. &#8220;work is said to be done when the ... 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