{"id":47320,"date":"2019-08-28T11:31:15","date_gmt":"2019-08-28T06:01:15","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=47320"},"modified":"2020-11-27T12:21:02","modified_gmt":"2020-11-27T06:51:02","slug":"plus-one-physics-chapter-wise-questions-answers-chapter-6","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/plus-one-physics-chapter-wise-questions-answers-chapter-6\/","title":{"rendered":"Plus One Physics Chapter Wise Questions and Answers Chapter 6 Work, Energy and Power"},"content":{"rendered":"

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 6 Work, Energy and Power<\/h2>\n

Plus One Physics Work, Energy and Power One Mark Questions and Answers<\/h3>\n

Question 1.
\nFind the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
\nAnswer:
\nkilowatt, unit of power
\n1 calorie = 4.2 joule
\n1 electron volt = 1.6 \u00d7 10-19<\/sup>J.<\/p>\n

Question 2.
\nWhat is the work done by the tension in the string of simple pendulum?
\nAnswer:
\nZero<\/p>\n

Question 3.
\nWhen is the exchange of energy is maximum during an elastic collision?
\nAnswer:
\nWhen mass of two colliding bodies are same, there will be maximum exchange of energy.<\/p>\n

Question 4.
\nIn atom, an electron is revolving around the nucleus. What is the work done?
\nAnswer:
\nWork done is zero because work done by centripetal force is zero.<\/p>\n

Question 5.
\nWhat is the type of collision when macroscopic particles collide?
\nAnswer:
\nPerfectly inelastic collision.<\/p>\n

Question 6.
\nName the parameter which is a measure of degree of elasticity of a body.
\nAnswer:
\nCoefficient of restitution.<\/p>\n

Question 7.
\nWhat is the source of kinetic energy for falling raindrops?
\nAnswer:
\nGravitational potential energy.<\/p>\n

Plus One Physics Work, Energy and Power Two Mark Questions and Answers<\/h3>\n

Question 1.
\nThe law of conversation of energy states that energy can neither be created nor be destroyed but can only change from one form into another. A bus and a car, moving with the same kinetic energy are brought to rest by applying an equal retardation force by the breaking systems. Which one will come to rest at a shorter distance? Give the reason behind your answer.
\nAnswer:
\nChange in K.E. = Force \u00d7 Displacement
\n1\/2 mv2<\/sup> = F \u00d7 S
\nie. KE \u03b1 s
\nKEcar<\/sub> \u03b1 Scar<\/sub> _____(1)
\nKEbus<\/sub> \u03b1 Sbus<\/sub> ______(2)
\n\"Plus
\nie. Scar<\/sub> = Sbus<\/sub>
\nBoth will travel equal distance.<\/p>\n

Question 2.
\nA body constrained to move along the Z-axis of a coordinate system is subjected to a constant force \\(\\bar{F}=(\\hat{i}+2 \\hat{\\jmath}+3 \\hat{k}) N\\)<\/p>\n

    \n
  1. What is the magnitude of force along z-direction?<\/li>\n
  2. What is the work done by this force in moving the body over a distance of 4m along the z-axis?<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. 3N<\/li>\n
    2. Work done = Force \u00d7 Displacement = 3 \u00d7 4 = 12J.<\/li>\n<\/ol>\n

      Question 3.
      \nMatch the following
      \n\"Plus
      \nAnswer:
      \nCollision of two balls – inelastic – TE and momentum Collision of two molecules – elastic – KE, TE, and momentum.<\/p>\n

      Plus One Physics Work, Energy and Power Three Mark Questions and Answers<\/h3>\n

      Question 1.
      \nA car of mass 1000kg moving with a speed 18mk\/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25 \u00d7 103<\/sup>N\/m<\/p>\n

        \n
      1. What do you mean by Spring constant.<\/li>\n
      2. What is the maximum compression of the spring?<\/li>\n<\/ol>\n

        Answer:
        \n1. Spring constant is the force required to stretch the spring by a unit distance.<\/p>\n

        2. \\(\\frac{1}{2}\\)mv2<\/sup> = \\(\\frac{1}{2}\\)kx2<\/sup>, 18km\/h = 5m\/s.
        \n\"Plus
        \nx = 2m.<\/p>\n

        Question 2.
        \nA man tries to lift a mass 200kg with a force 100N<\/p>\n

          \n
        1. Is he doing work? Explain.<\/li>\n
        2. If yes, find the amount of work done If No, find the force required to lift it.<\/li>\n
        3. If it is lifted to 2m in 10 seconds, find his power.<\/li>\n<\/ol>\n

          Answer:<\/p>\n

            \n
          1. No work is done, as there is no displacement, 100N force is insufficient to raise 200kg.<\/li>\n
          2. Force required to lift 200 kg = 200 \u00d7 9.8 = 1960N<\/li>\n
          3. Power = \\(\\frac{m g h}{t}=\\frac{200 \\times 9.8 \\times 2}{10}\\) = 392W.<\/li>\n<\/ol>\n

            Question 3.
            \nTwo cricket balls are colliding each other.<\/p>\n

              \n
            1. Name the collision<\/li>\n
            2. Say whether law of conservation of Kinetic Energy hold good in this case. Why?<\/li>\n
            3. State and prove the other conservation law applicable here.<\/li>\n<\/ol>\n

              Answer:<\/p>\n

                \n
              1. Inelastic collision<\/li>\n
              2. No, Total KE before collision is not equal to total KE after collision.<\/li>\n
              3. Proof and statement of law of conservation of momentum.<\/li>\n<\/ol>\n

                Plus One Physics Work, Energy and Power Four Mark Questions and Answers<\/h3>\n

                Question 1.
                \nTwo cars A and B travelling with speeds 20m\/s and 10m\/s respectively applies breaks,.so that A comes to rest in 15 second and B in 7.5s<\/p>\n

                  \n
                1. From the graph determine which of the two cars travelled further after brakes were applied and by how much distance it travelled?<\/li>\n
                2. Draw the velocity-time graph of A and B in the same graph.<\/li>\n
                3. In the above process the wear and tear of which the car gets affected more?<\/li>\n<\/ol>\n

                  Answer:
                  \n1. The area of velocity time graph gives displacement distance travelled by car A, SA<\/sub> = 1\/2 \u00d7 20 \u00d7 15 = 150 m
                  \ndistance travelled by the car B, SB<\/sub> = 1\/2 \u00d7 10 \u00d7 7.5 = 37.5.<\/p>\n

                  2.
                  \n\"Plus<\/p>\n

                  3. Wear and tear gets affected more for car A.<\/p>\n

                  Question 2.
                  \nA sphere of mass m is moving with a velocity u and makes a head-on collision with another identical mass which is at rest. It is observed that the stationary mass starts moving with a lesser velocity than u, after the collision.<\/p>\n

                    \n
                  1. Which physical quantity is conserved here?<\/li>\n
                  2. Define coefficient of restitution.<\/li>\n
                  3. Determine the ratio of the velocities of the two spheres after elastic collision if \u2018e\u2019 is the coefficient of restitution.<\/li>\n<\/ol>\n

                    Answer:
                    \n1. conservation of linear momentum.<\/p>\n

                    2. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.<\/p>\n

                    3. Coefficient of restitution,
                    \ne = \\(\\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\\)
                    \nin this cos u1<\/sub> = u, u2<\/sub> = 0, v1<\/sub> = 0, v2<\/sub> = u
                    \n\u2234 e = \\(\\frac{u-0}{u-0}\\)
                    \ne = 1.<\/p>\n

                    Question 3.
                    \nFrom the table given below<\/p>\n

                      \n
                    1. Draw the force-displacement curve<\/li>\n
                    2. Analyse the graph & find the type of force involved<\/li>\n
                    3. Estimate the workdone<\/li>\n<\/ol>\n

                      \"Plus
                      \nAnswer:
                      \n1.
                      \n\"Plus<\/p>\n

                      2. Workdone by a variable force.<\/p>\n

                      3. Workdone = Area of the graph
                      \n= \\(\\frac{1}{2}\\)bh
                      \n= \\(\\frac{1}{2}\\)5 \u00d7 10 = 25J.<\/p>\n

                      Question 4.
                      \nRaju increased the speed of moving mass \u201850 kg\u2019 from 2 m\/s to 4m\/s.<\/p>\n

                        \n
                      1. How much force will be required, if velocity change takes place within 0.2 sec?<\/li>\n
                      2. How much work is done by Raju?<\/li>\n<\/ol>\n

                        Answer:
                        \n1. F = mass \u00d7 acceleration
                        \n= 50 \u00d7 \\(\\frac{(4-2)}{0.2}\\)
                        \n= 500N.<\/p>\n

                        2. w = \\(\\frac{1}{2}\\)mv2<\/sup> – \\(\\frac{1}{2}\\)mu2<\/sup>.
                        \n=\\(\\frac{1}{2}\\)50 (42<\/sup> – 22<\/sup>)
                        \n= 300 J.<\/p>\n

                        Plus One Physics Work, Energy and Power Five Mark Questions and Answers<\/h3>\n

                        Question 1.
                        \nA car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. (Assume that the truck is heavier than the car)<\/p>\n

                          \n
                        1. Which one will have greater momentum?<\/li>\n
                        2. Write the relationship between kinetic energy and linear momentum.<\/li>\n
                        3. If the mass of truck is 100 times greater than that of the car, find the ratio between their velocities.<\/li>\n<\/ol>\n

                          Answer:
                          \n1. Kinetic energy, of car, K.Ec<\/sub> = \\(\\frac{P_{c}^{2}}{2 m_{c}}\\)
                          \nKinetic energy of truck,
                          \n\"Plus
                          \nSince mc<\/sub> < mt<\/sub>
                          \nHence Pt<\/sub> > Pc<\/sub>
                          \n\u2234 momentum of truck is greater than car.<\/p>\n

                          2. KE = P2<\/sup>\/2m.<\/p>\n

                          3. KEc<\/sub> = KEt<\/sub>
                          \n1\/2mc<\/sub> Vc<\/sub>2<\/sup> = 1\/2 x mt<\/sub>Vt<\/sub>2<\/sup>
                          \nBut mt<\/sub> = 100mc<\/sub>
                          \n\\(V_{t}^{2}=\\frac{V_{c}^{2}}{100}\\)
                          \nVelocity of truck, Vt = \\(\\frac{V_{c}}{10}\\)
                          \nratio of velocity, 10Vt<\/sub> = Vc<\/sub>
                          \n10:1.<\/p>\n

                          Question 2.
                          \nRaju dropped a rubber ball of mass m from a height h to the ground. He observed that the ball rebounds vertically and along the same line to a height h1<\/sub>, which is less than h.<\/p>\n

                            \n
                          1. Is it an elastic or inelastic collision?<\/li>\n
                          2. Find the velocity with which it strikes the ground?<\/li>\n
                          3. If it is replaced by a solid aluminium ball, then what happens to the height of rebound?<\/li>\n
                          4. If the rubber ball is allowed to fall on a spring placed on the ground then what change will Raju notice in the height of rebound?<\/li>\n<\/ol>\n

                            Answer:
                            \n1. Inelastic collision.<\/p>\n

                            2. The velocity with which ball strikes the ground,
                            \nv2<\/sup> = u2<\/sup> + 2as
                            \nv2<\/sup> = 0 + 2g \u00d7 h
                            \nv = \\(\\sqrt{2 g h}\\).<\/p>\n

                            3. Height of rebound decreases.<\/p>\n

                            4. Height of rebound depends on the state of potential energy stored in the spring. If ball falls on a compressed spring, the height of rebound increases due to potential energy given by the spring to bail.<\/p>\n

                            Question 3.
                            \nA man tries to lift a mass 200kg with a force 100N.<\/p>\n

                              \n
                            1. Is he doing work? Explain.<\/li>\n
                            2. If it is lifted to 2m in 10s, find the power.<\/li>\n
                            3. Show that total mechanical energy is conserved fora freely falling body.<\/li>\n<\/ol>\n

                              Answer:
                              \n1. No. Force required to lift the body is 2000N (w = mg = 200 \u00d7 10). But the applied force is 100N. Hence there is no displacement due to this applied force.<\/p>\n

                              2. Power P = \\(\\frac{w}{t}=\\frac{m g h}{t}=\\frac{200 \\times 10 \\times 2}{10}\\) = 400 watt.<\/p>\n

                              3.
                              \n\"Plus
                              \nConsider a body of mass \u2018m\u2019 at a height h from the ground.
                              \nTotal energy at point A<\/span>
                              \nPotential energy at A,
                              \nPE = mgh
                              \nKinetic energy, KE = \\(\\frac{1}{2}\\) mv2<\/sup> = 0
                              \n(since the body at rest, v = 0).
                              \n\u2234 Total mechanical energy = PE + KE
                              \n= mgh + 0 = mgh
                              \nTotal energy at the point B<\/span>
                              \nThe body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
                              \nv2<\/sup> = u2<\/sup> + 2as
                              \nv2<\/sup> = 0 + 2 gx
                              \n\u2234 KE at B, = \\(\\frac{1}{2}\\) mv2<\/sup>
                              \n= \\(\\frac{1}{2}\\) m2gx
                              \n= mgx
                              \nP.E. at B, = mg(h – x)
                              \nTotal mechanical energy = PE + KE
                              \n= mg(h – x) + mgx
                              \n= mgh
                              \nTotal energy at C<\/span>
                              \nVelocity at C can be found using the formula
                              \nv2<\/sup> = u2<\/sup> + 2as
                              \nv2<\/sup> = 0 + 2 gh
                              \n\u2234 KE at C, = \\(\\frac{1}{2}\\)mv2<\/sup>
                              \n= \\(\\frac{1}{2}\\)m2gh
                              \n= mgh
                              \nP.E. at C = 0
                              \nTotal energy = PE + KE = 0 + mgh = mgh.<\/p>\n

                              Question 4.
                              \nAn elevator of total mass 1800kg is moving up with a constant speed of 2m\/s. A frictional force of 400N acts on this motion.<\/p>\n

                                \n
                              1. The direction of frictional force is______\n