{"id":244,"date":"2023-03-02T13:21:58","date_gmt":"2023-03-02T13:21:58","guid":{"rendered":"https:\/\/www.vartmaaninstitutesirsa.com\/?page_id=244"},"modified":"2023-03-31T12:56:07","modified_gmt":"2023-03-31T12:56:07","slug":"unit-iv-work-energy-and-power","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/unit-iv-work-energy-and-power\/","title":{"rendered":"Unit IV: Work Energy and Power"},"content":{"rendered":"\r\n

Chapter\u20136: Work Energy and Power<\/strong><\/h1>\r\n\r\n\r\n\r\n

Work Energy and Power :<\/strong>Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power.<\/p>\r\n\r\n\r\n\r\n

Notion of potential energy, potential energy of a spring, conservative forces: conservation of mechanical energy (kinetic and potential energies); non-conservative forces: motion in a vertical circle; elastic and inelastic collisions in one and two dimensions<\/p>\r\n\r\n\r\n\r\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/b>UNIT 4 Work <\/b>Energy and\u00a0<\/b>Power<\/b><\/h2>\r\n\r\n\r\n\r\n
    \r\n
  • WORK<\/b><\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n

    Work is said to be done when a force applied on a body displace the body through a distance in the direction of force.<\/span><\/p>\r\n\r\n\r\n\r\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b>Let us consider that force F displace the body from point A to point B through a distance S. than work done to displace the body is \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span> W=<\/span>F<\/span>.<\/span>S<\/span><\/p>\r\n\r\n\r\n\r\n

    When force act at an angle \u03b8 with horizontal:-<\/b><\/p>\r\n\r\n\r\n\r\n

    Suppose a force F is applied on a body at an angle \u03b8 with the horizontal and displace the body from point A to B through a distance S.<\/span> Now resolving F into components<\/span><\/p>\r\n\r\n\r\n\r\n

      \r\n
    1. F Cos\u03b8<\/span> in direction of displacement.<\/span><\/li>\r\n\r\n\r\n\r\n
    2. F Sin\u03b8<\/span> in perpendicular direction of displacement.<\/span><\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

      Clearly here the motion is due to <\/span>Fcos\u03b8<\/span> so <\/span>W=FS cos\u03b8<\/span>\u00a0\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or \u00a0 <\/span>W=<\/span>F<\/span>.<\/span>S<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

      Thus work done by a force is equal to dot product of force and displacement of the body.<\/span><\/p>\r\n\r\n\r\n\r\n

      Dimensions and Unit of Work:-\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/p>\r\n\r\n\r\n\r\n

      \u00a0<\/b>As \u00a0 \u00a0 work=force \u00d7 displacement<\/span> \u00a0\u00a0\u00a0So\u00a0 [W] =[MLT<\/span>-2<\/span>][L] \u00a0 \u00a0 <\/span>Or \u00a0 \u00a0 [W]= [ML<\/b>2<\/b>T<\/b>-2<\/b>]<\/b><\/p>\r\n\r\n\r\n\r\n

      Unit of work <\/b>There are of two types of unit of work\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

        \r\n
      1. Absolute unit of work:-<\/b><\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(i) <\/span>In SI the absolute unit of work is Joule.<\/b><\/p>\r\n\r\n\r\n\r\n

        If\u00a0 <\/span>F=1N and S=1 metre<\/span> than <\/span>1 Joule= 1N\u00d71m<\/span><\/p>\r\n\r\n\r\n\r\n

        Thus, work done is said to be one joule if one Newton force displaces the body through one meter in its own direction.<\/span><\/p>\r\n\r\n\r\n\r\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii<\/span>) In c,g,s system the absolute unit of work is erg.<\/b><\/p>\r\n\r\n\r\n\r\n

        If <\/span>F = 1 dyne\u00a0 and \u00a0 S = 1cm\u00a0 then\u00a0 1 Erg = 1 dyne \u00d7 1 cm.<\/span><\/p>\r\n\r\n\r\n\r\n

        Hence work done is said to be one erg if one dyne force displace a body to one metre distance in its own direction. The relation between one joule and one erg is<\/span><\/p>\r\n\r\n\r\n\r\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>1 J=10<\/b>7<\/b> erg<\/b><\/p>\r\n\r\n\r\n\r\n

          \r\n
        1. Gravitational unit of work:-<\/b><\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

          (i)\u00a0 In SI the gravitational unit of work is kilogram metre<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>1 kilogram metre \u00a0 = 1kg wt \u00d7 1m=9.8N\u00d71m<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span>1kg m=9.8\u00a0 J<\/span><\/p>\r\n\r\n\r\n\r\n

          Thus, work done is said to be one kg m if one kg wt force displaces the body through one meter distance in its own direction.\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

          (ii)\u00a0 In cgs the gravitational unit of work is Gram centimeter\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>1 Gram centimeter = 1g wt \u00d7 1 cm= 980 dyne\u00d7 1 cm.<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>1g cm=980 erg\u00a0<\/b><\/p>\r\n\r\n\r\n\r\n

          Thus, work done is said to be one g cm if one g wt force displaces the body through one centimeter distance in its own direction.<\/span><\/p>\r\n\r\n\r\n\r\n

          Question 1 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the<\/b>
          <\/span>(a) Work done by the applied force in 10 s<\/b>
          <\/span>(b) Work done by friction in 10 s<\/b>
          <\/span>(c) Work done by the net force on the body in 10 s<\/b>
          <\/span>(d) Change in kinetic energy of the body in 10 s and interpret your results.<\/b>
          <\/span>Answer:\u00a0<\/b>(a) We know that <\/span>frictional force = <\/span>k<\/span>\u00a0<\/span> normal reaction<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Or \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span>= 0.1 <\/span> 2 kg wt = 0.1 <\/span> 2 <\/span> 9.8 N = 1.96 N<\/span>or <\/span>net effective force = (7 \u2013 1.96) N = 5.04 N<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

          \u00a0so \u00a0 <\/span>acceleration =<\/span>F<\/span>m<\/span>=<\/span>5.04<\/span>2<\/span> m<\/span>s<\/span>–<\/span>2<\/span>\u00a0= 2.52 m<\/span>s<\/span>–<\/span>2<\/span>and distance traveled\u00a0 , <\/span>S=<\/span>1<\/span>2<\/span>a<\/span>t<\/span>2<\/span>=<\/span>1<\/span>2<\/span>\u00d7 2.52 \u00d7 10 \u00d7 10 = 126 m<\/span>so work done by applied force <\/span>=FS= 7 \u00d7 126 J = 882 J<\/span>
          <\/span>(b) Work done by friction = <\/span>fS=1.96 \u00d7 126 = -246.96 J<\/span>(c) Work done by net force <\/span>= 5.04 \u00d7 126 = 635.04 J<\/span>(d) Change in the kinetic energy of the body = work done by the net force in 10 seconds = 635.04 J\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

          Question 2. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km\/h on a friction less track. After a while, sand starts leaking out of a hole on the trolley\u2019s floor at the rate of 0.05 kg s<\/b>-1<\/b>. What is the speed of the trolley after the entire sand bag is empty?<\/b>
          <\/span>Answer:\u00a0<\/b>\u00a0The system of trolley and sandbag is moving with a uniform speed. Clearly, the system is not being acted upon by an external force. If the sand leaks out, even then no external force acts. So there shall be no change in the speed of the trolley.<\/span><\/p>\r\n\r\n\r\n\r\n

            \r\n
          1. Nature of work done:-<\/b><\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

            1 If <\/b>\u03b8 <9<\/span>0<\/span>0<\/span> than Cos\u03b8 is +ve \u00a0 so\u00a0 W =FS (+ve)<\/span><\/p>\r\n\r\n\r\n\r\n

            (a) When a body is lifted, the work done by the lifting force is positive.<\/span><\/p>\r\n\r\n\r\n\r\n

            (b) When a spring is stretched, work done by the stretching force is positive.<\/span><\/p>\r\n\r\n\r\n\r\n

            2 If <\/b>\u03b8=9<\/span>0<\/span>0<\/span> than =FScos9<\/span>0<\/span>0<\/span> =0<\/span><\/p>\r\n\r\n\r\n\r\n

            (a)\u00a0 A person holding a heavy stone at rest is doing no work. I.e. S=0\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n

            (b) A coolie carrying a suitcase on his head is doing no work. I.e. \u03b8=90<\/span>0<\/span><\/p>\r\n\r\n\r\n\r\n

            3 If <\/b>>9<\/span>0<\/span>0<\/span>\u00a0 than W = -FS (-ve)<\/span><\/p>\r\n\r\n\r\n\r\n

            (a)When break are applied on a moving vehicle, the work done by the breaking force is negative.<\/span><\/p>\r\n\r\n\r\n\r\n

            (b) Work done by the frictional force is negative.<\/span><\/p>\r\n\r\n\r\n\r\n

            Question\u00a0 3. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:<\/b>
            <\/span>(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,<\/b>
            <\/span>(b) Work done by gravitational force in the above case,<\/b>
            <\/span>(c) Work done by friction on a body sliding down an inclined plane,<\/b>
            <\/span>(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,<\/b>
            <\/span>(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.<\/b>
            <\/span>Answer:<\/b>\u00a0Work done<\/span>, W = F.S = FS cos \u03b8<\/span>
            <\/span>(a) Work done \u2018positive\u2019, because force is acting in the direction of displacement i.e., \u03b8 = 0\u00b0.<\/span>
            <\/span>(b) Work done is negative, because force is acting against the displacement i.e., \u03b8 = 180\u00b0.<\/span>
            <\/span>(c) Work done is negative, because force of friction is acting against the displacement i.e., \u03b8=180\u00b0.<\/span>
            <\/span>(d) Work done is positive, because body moves in the direction of applied force i.e., \u03b8= 0\u00b0.<\/span>
            <\/span>(e) Work done is negative, because the resistive force of air opposes the motion i.e., \u03b8 = 180\u00b0.<\/span><\/p>\r\n\r\n\r\n\r\n

            Question 4 : In Fig.(i), the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?<\/b>
            <\/span>Answer:<\/b> In Fig. (i), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span>\u03b8 = 90\u00b0. \u00a0 W = F<\/span>s<\/span>\u00a0cos 90\u00b0 = zero<\/span>In Fig. (ii), force is applied along the horizontal and the distance moved is also along the horizontal. Therefore, \u03b8 = 0\u00b0.<\/span>
            <\/span>W = F<\/span>s<\/span>\u00a0cos \u03b8 = mg \u00d7 s cos 0\u00b0W = 15 \u00d7 9.8 \u00d7 2 \u00d7 1 = 294 joule.<\/span><\/p>\r\n\r\n\r\n\r\n


            <\/span>Thus, work done in (ii) case is greater.<\/span><\/p>\r\n\r\n\r\n\r\n