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NCERT Exemplar · Class 9 Science Work and Energy

28 questions · 28 still being checked

Long Answer Questions 21–28 (part 3 of 3)

  1. Exercise 21

    A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

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    NCERT’s answer
    \(\displaystyle p_{1}\) = \(\displaystyle m_{1}\)\(\displaystyle v_{1}\) \(\displaystyle p_{2}\) = \(\displaystyle m_{2}\) \(\displaystyle v_{2}\) But \(\displaystyle p_{1}\) = \(\displaystyle p_{2}\) or \(\displaystyle m_{1}\)\(\displaystyle v_{1}\) = \(\displaystyle m_{2}\)\(\displaystyle v_{2}\) If \(\displaystyle m_{1}\)< \(\displaystyle m_{2}\) then \(\displaystyle v_{1}\)> \(\displaystyle v_{2}\) (K.E)$\displaystyle 1$ = m v (K.E)$\displaystyle 2$ = m v (K.E)$\displaystyle 1$= m v v $\displaystyle 1$ $\displaystyle 1$ $\displaystyle 1$ ( ) p v = (K.E)$\displaystyle 2$ = v v $\displaystyle 1$ (m ) = p v = = (K.E) (K.E) \(\displaystyle p_{2}\) \(\displaystyle v_{1}\) \(\displaystyle p_{2}\) v v v = But \(\displaystyle v_{1}\)> \(\displaystyle v_{2}\) Therefore, (K.E.)$\displaystyle 1$ > (K.E)$\displaystyle 2$
    Equal momentum for both bodies. \[p_L = p_H = p = mv \] \[KE = \frac{p^{2}}{2m} \] \[\frac{KE_L}{KE_H} = \frac{m_H}{m_L} \] Since \(\displaystyle m_H > m_L\), this ratio exceeds 1.Answer: \(\displaystyle KE_L : KE_H = m_H : m_L\); the lighter object has the larger kinetic energy.

  2. Exercise 22

    An automobile engine propels a 1000\displaystyle 1000 kg car (A) along a levelled road at a speed of 36\displaystyle 36 km h−1\displaystyle h^{-1}. Find the power if the opposing frictional force is 100\displaystyle 100 N. Now, suppose after travelling a distance of 200\displaystyle 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

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    NCERT’s answer
    \(\displaystyle m_{(A)}\)= \(\displaystyle m_{(B)}\) =$\displaystyle 1000$ kg. v = $\displaystyle 36$ km/h =$\displaystyle 10$ m/s Frictional force = $\displaystyle 100$ N Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force Power = F v Force × distance = . time = $\displaystyle 100$ N × $\displaystyle 10$ m/s = $\displaystyle 1000$ W After collision \(\displaystyle m_{A}\) \(\displaystyle u_{A}\) + \(\displaystyle m_{B}\) \(\displaystyle u_{B}\) = \(\displaystyle m_{A}\) \(\displaystyle v_{A}\) + \(\displaystyle m_{B}\) \(\displaystyle v_{B}\). $\displaystyle 1000$ ×$\displaystyle 10$ +$\displaystyle 1000$ × $\displaystyle 0$ = $\displaystyle 1000$ × $\displaystyle 0$ + $\displaystyle 1000$ × \(\displaystyle v_{B}\) \(\displaystyle v_{B}\) = $\displaystyle 10$ m \(\displaystyle s^{-1}\)
    \[v = 36\text{ km h}^{-1} = 10\text{ m s}^{-1} \]At constant speed the engine force balances friction, so \[P = Fv = (100\text{ N})(10\text{ m s}^{-1}) = 1000\text{ W} = 1\text{ kW} \]Car A, still moving at \(\displaystyle 10\text{ m s}^{-1}\), strikes stationary car B and stops; momentum is conserved. \[m_Av + m_B(0) = m_A(0) + m_Bv' \] \[(1000\text{ kg})(10\text{ m s}^{-1}) = (1000\text{ kg})\,v' \] \[v' = 10\text{ m s}^{-1} = 36\text{ km h}^{-1} \]Answer: \(\displaystyle P = 1\text{ kW}\); car B moves off at \(\displaystyle v' = 10\text{ m s}^{-1} = 36\text{ km h}^{-1}\).

  3. Exercise 23

    A girl having mass of 35\displaystyle 35 kg sits on a trolley of mass 5\displaystyle 5 kg. The trolley is given an initial velocity of 4\displaystyle 4 m s−1\displaystyle s^{-1} by applying a force. The trolley comes to rest after traversing a distance of 16\displaystyle 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

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    NCERT’s answer
    u = $\displaystyle 4$ m \(\displaystyle s^{-1}\) , v =o, s = $\displaystyle 16$ m a = v u − = − = − × - $\displaystyle 1$ m s 2s Force = m a = $\displaystyle 40$ × ( ) − = - $\displaystyle 20$ N Work done on the trolley = $\displaystyle 20$ N ×$\displaystyle 16$ m = $\displaystyle 320$ J Work done by the girl = $\displaystyle 0$ J.
    Total mass \(\displaystyle M = 35+5 = 40\text{ kg}\), \(\displaystyle u = 4\text{ m s}^{-1}\), \(\displaystyle v = 0\), \(\displaystyle s = 16\text{ m}\). By the work-energy theorem, \[W = \Delta KE = \frac{1}{2}Mv^{2} - \frac{1}{2}Mu^{2} \] \[W = 0 - \frac{1}{2}(40\text{ kg})(4\text{ m s}^{-1})^{2} = -320\text{ J} \] The girl moves with the trolley as part of the $\displaystyle 40$ kg body; she does no work of her own -- the retarding work is done by friction between the trolley and the road.Answer: work done on the trolley \(\displaystyle = 320\text{ J}\) (magnitude, by friction); work done by the girl \(\displaystyle = 0\text{ J}\).

  4. Exercise 24

    Four men lift a 250\displaystyle 250 kg box to a height of 1\displaystyle 1 m and hold it without raising or
    lowering it. (a) How much work is done by the men in lifting the box?
    (b)
    How much work do they do in just holding it? (c) Why do they get tired
    while holding it? (g = 10\displaystyle 10 m s−2\displaystyle s^{-2})

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    NCERT’s answer
    (a)
    F = $\displaystyle 250$ kg × g (g =$\displaystyle 10$ m \(\displaystyle s^{-2}\)) = $\displaystyle 2500$ N s = $\displaystyle 1$ m W = F.s = $\displaystyle 2500$ N m = $\displaystyle 2500$ J (b) zero; as the box does not move at all, while holding it. (c) In order to hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force, muscular effort is involved. So they feel tired.
    \[W_{\text{lift}} = mgh = (250\ \text{kg})(10\ \text{m s}^{-2})(1\ \text{m}) = 2500\ \text{J} \]Holding it stationary means displacement \(\displaystyle =0\), so \[W_{\text{hold}} = Fd\cos\theta = F\times0 = 0\ \text{J} \]The muscles keep contracting to supply an upward force equal to the weight; that energy is lost as heat, even though no mechanical work is done on the box.Answer: (a) \(\displaystyle 2500\ \text{J}\) (b) \(\displaystyle 0\ \text{J}\) (c) muscular effort dissipates energy as heat despite zero mechanical work.

  5. Exercise 25

    What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20\displaystyle 20 m high. 2000\displaystyle 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10\displaystyle 10 m s−2\displaystyle s^{-2})

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    NCERT’s answer
    Power is the rate of doing work. Kilowatt is the unit of power and kilowatt hour is the unit of energy. h = $\displaystyle 20$ m, and mass = $\displaystyle 2000$ × \(\displaystyle 10^{3}\) kg = $\displaystyle 2$ × \(\displaystyle 10^{6}\) kg Power = m g h t $\displaystyle 2$×$\displaystyle 10$ ×$\displaystyle 10$ = × $\displaystyle 20$ W = × $\displaystyle 10$ W = × $\displaystyle 10$ W
    Power is the rate of doing work, \[P = \frac{W}{t} \] A kilowatt measures this rate; a kilowatt hour measures energy -- the work done when $\displaystyle 1$ kW runs for $\displaystyle 1$ hour, \[1\ \text{kWh} = 1000\ \text{W}\times3600\ \text{s} = 3.6\times10^{6}\ \text{J} \] Mass of water \(\displaystyle m = 2000\ \text{tonnes} = 2\times10^{6}\ \text{kg}\) falls height \(\displaystyle h = 20\ \text{m}\) in \(\displaystyle t = 60\ \text{s}\): \[E = mgh = (2\times10^{6}\ \text{kg})(10\ \text{m s}^{-2})(20\ \text{m}) = 4\times10^{8}\ \text{J} \] \[P = \frac{E}{t} = \frac{4\times10^{8}\ \text{J}}{60\ \text{s}} = 6.67\times10^{6}\ \text{W} \] Answer: \(\displaystyle P \approx 6.67\times10^{6}\ \text{W}\) (\(\displaystyle \approx 6.67\) MW).

  6. Exercise 26

    How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100\displaystyle 100 W, be able to lift at constant speed of 1\displaystyle 1 m s−1\displaystyle s^{-1} vertically? (g = 10\displaystyle 10 m s−2\displaystyle s^{-2})

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    NCERT’s answer
    Power = work done or energy = time m g h h m. g. t t = ( ) Here h t = speed Therefore, m = g power = =10kg × speed $\displaystyle 10$×$\displaystyle 1$
    Power is work done per unit time; lifting at constant speed means the applied force equals the body's weight, so power grows in step with speed.\[P = 100\ \text{W}, \qquad v = 1\ \text{m s}^{-1}, \qquad g = 10\ \text{m s}^{-2} \] \[P = Fv = (mg)v \] \[m = \frac{P}{gv} = \frac{100\ \text{W}}{10\ \text{m s}^{-2} \times 1\ \text{m s}^{-1}} \] \[m = 10\ \text{kg} \]Answer: \(\displaystyle P = mgv\), directly proportional to speed; the man can lift $\displaystyle 10$ kg.

  7. Exercise 27

    Define watt. Express kilowatt in terms of joule per second. A 150\displaystyle 150 kg car engine develops 500\displaystyle 500 W for each kg. What force does it exert in moving the car at a speed of 20\displaystyle 20 m s−1\displaystyle s^{-1}?

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    NCERT’s answer
    One watt is the power of an agent which does work at the rate of 1J \(\displaystyle s^{-1}\) $\displaystyle 1$ kilowatt = $\displaystyle 1000$ J \(\displaystyle s^{-1}\) Total Power = $\displaystyle 150$ × $\displaystyle 500$ = $\displaystyle 7.5$ × \(\displaystyle 10^{4}\) W Force = Power velocity = $\displaystyle 7.5$ ×$\displaystyle 10$ = $\displaystyle 3.75$ ×$\displaystyle 10$ N Force = $\displaystyle 3750$ N.
    One watt is the power when one joule of work is done in one second.\[1\ \text{W} = 1\ \text{J s}^{-1}, \qquad 1\ \text{kW} = 1000\ \text{W} = 1000\ \text{J s}^{-1} \]Total power developed by the engine:\[P = 150\ \text{kg} \times 500\ \text{W kg}^{-1} = 75000\ \text{W} \]\[P = Fv \] \[F = \frac{P}{v} = \frac{75000\ \text{W}}{20\ \text{m s}^{-1}} \] \[F = 3750\ \text{N} \]Answer: \(\displaystyle 1\ \text{W} = 1\ \text{J s}^{-1}\), \(\displaystyle 1\ \text{kW} = 1000\ \text{J s}^{-1}\); the engine exerts $\displaystyle 3750$ N.

  8. Exercise 28

    Compare the power at which each of the following is moving upwards
    against the force of gravity? (given g = 10\displaystyle 10 m s−2\displaystyle s^{-2})
    (i)
    a butterfly of mass 1.0\displaystyle 1.0 g that flies upward at a rate of 0.5\displaystyle 0.5 m s−1\displaystyle s^{-1}.
    (ii)
    a 250\displaystyle 250 g squirrel climbing up on a tree at a rate of 0.5\displaystyle 0.5 m s−1\displaystyle s^{-1}.

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    NCERT’s answer
    (i)
    Power = mg × velocity, g = $\displaystyle 10$ m \(\displaystyle s^{-2}\) = ×$\displaystyle 10$× 0.5W $\displaystyle 1000$ = $\displaystyle 0.5$ -$\displaystyle 3$ W =$\displaystyle 5$×$\displaystyle 10$ W (ii) Power = $\displaystyle 250$ ×$\displaystyle 10$× $\displaystyle 0.5$ W $\displaystyle 1000$ = . × × $\displaystyle 0$ $\displaystyle 5$ = $\displaystyle 1.25$ W Hence, the power with which the squirrel is climbing is much higher than that of a butterfly flying.
    Both climb at the same speed; only their weight differs, so power scales with mass.\[m_1 = 1.0\ \text{g} = 1.0\times10^{-3}\ \text{kg}, \qquad v = 0.5\ \text{m s}^{-1}, \qquad g = 10\ \text{m s}^{-2} \] \[P_{\text{butterfly}} = m_1 g v = (1.0\times10^{-3})(10)(0.5) \] \[P_{\text{butterfly}} = 5\times10^{-3}\ \text{W} \]\[m_2 = 250\ \text{g} = 0.25\ \text{kg} \] \[P_{\text{squirrel}} = m_2 g v = (0.25)(10)(0.5) \] \[P_{\text{squirrel}} = 1.25\ \text{W} \]\[\frac{P_{\text{squirrel}}}{P_{\text{butterfly}}} = \frac{1.25}{5\times10^{-3}} = 250 \]Answer: the squirrel develops $\displaystyle 250$ times the butterfly's power.