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NCERT Exemplar · Class 9 Science Work and Energy

28 questions · 28 still being checked

Short Answer Questions 10–20 (part 2 of 3)

  1. Exercise 10

    A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

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    NCERT’s answer
    Initial velocity = v, then v′ = $\displaystyle 3$ v Initial kinetic energy = $\displaystyle 2$ m \(\displaystyle v^{2}\) Final kinetic energy (K.E.) = m v m v m v ′$\displaystyle 2$ = ($\displaystyle 3$ ) = $\displaystyle 9$ ( ) (K.E) initial : (K.E) final = $\displaystyle 1$:$\displaystyle 9$
    \[KE = \frac{1}{2}mv^2 \] \[KE_1 = \frac{1}{2}mv^2, \quad KE_2 = \frac{1}{2}m(3v)^2 = 9\left(\frac{1}{2}mv^2\right) \] \[\frac{KE_1}{KE_2} = \frac{1}{9} \]Tripling the speed makes the kinetic energy nine times as large.Answer: \(\displaystyle KE_1 : KE_2 = 1:9\).

  2. Exercise 11

    Avinash can run with a speed of 8\displaystyle 8 m s−1\displaystyle s^{-1} against the frictional force of 10\displaystyle 10 N, and Kapil can move with a speed of 3\displaystyle 3 m s−1\displaystyle s^{-1} against the frictional force of 25\displaystyle 25 N. Who is more powerful and why?

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    NCERT’s answer
    Power of Avinash \(\displaystyle P_{A}\) = \(\displaystyle F_{A}\) . \(\displaystyle v_{A}\) = $\displaystyle 10$ × $\displaystyle 8$ = $\displaystyle 80$ W The power of Kapil \(\displaystyle P_{k}\) = \(\displaystyle F_{k}\) . \(\displaystyle v_{k}\) = $\displaystyle 25$ × $\displaystyle 3$ = $\displaystyle 75$ W So, Avinash is more powerful than Kapil.
    \[P = Fv \] \[P_{Avinash} = 10\,\text{N} \times 8\,\text{m s}^{-1} = 80\,\text{W} \] \[P_{Kapil} = 25\,\text{N} \times 3\,\text{m s}^{-1} = 75\,\text{W} \]At constant speed the applied force equals the opposing friction, so power is force times velocity.Answer: Avinash is more powerful -- \(\displaystyle 80\,\text{W} > 75\,\text{W}\).

  3. Exercise 12

    A boy is moving on a straight road against a frictional force of 5\displaystyle 5 N. After travelling a distance of 1.5\displaystyle 1.5 km he forgot the correct path at a round about (Fig. 11.1\displaystyle 11.1) of radius 100\displaystyle 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0\displaystyle 2.0 km. Calculate the work done by him. NCERT_Question_Class9_Science_Exemplar_Ch11_Q12

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    Friction is the only opposing force on the level road, so the work he does equals the work needed to overcome friction over the whole path, \(\displaystyle W = F \times d\). Circumference of the roundabout (Fig. $\displaystyle 11.1$): \[C = 2\pi r = 2 \times 3.14 \times 100\ \text{m} = 628\ \text{m} \] Distance for one and a half revolutions: \[d_{\text{circle}} = 1.5 \times 628\ \text{m} = 942\ \text{m} \] Total distance travelled: \[d = 1500\ \text{m} + 942\ \text{m} + 2000\ \text{m} = 4442\ \text{m} \] Work done by the boy: \[W = 5\ \text{N} \times 4442\ \text{m} = 22210\ \text{J} \]Answer: \(\displaystyle W = 22210\ \text{J} \approx 2.22\times10^{4}\ \text{J}\)NCERT prints: \(\displaystyle W = 5 \times [1500 + 200 + 2000] = 18500\ \text{J}\) — the $\displaystyle 200$ m stands in for the roundabout's diameter, not the path length of one and a half turns; work against friction is force times path length, and \(\displaystyle 1.5 \times 2\pi(100\ \text{m}) = 942\ \text{m}\) gives \(\displaystyle 5 \times 4442\ \text{m} = 22210\ \text{J}\).

  4. Exercise 13

    Can any object have mechanical energy even if its momentum is zero? Explain.

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    NCERT’s answer
    Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.
    \[p = mv \] \[v = 0 \;\Rightarrow\; p = 0, \quad KE = \frac{1}{2}mv^2 = 0 \] \[E = KE + PE = PE = mgh \]A book at rest on a shelf of height \(\displaystyle h\) has zero velocity, hence zero momentum, yet nonzero potential energy.Answer: Yes -- any stationary object above the ground has \(\displaystyle p=0\) but \(\displaystyle E = mgh \neq 0\).

  5. Exercise 14

    Can any object have momentum even if its mechanical energy is zero? Explain.

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    NCERT’s answer
    No. Since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.
    \[E = KE + PE = 0,\quad KE \ge 0,\ PE \ge 0 \;\Rightarrow\; KE = 0,\ PE = 0 \] \[KE = \tfrac{1}{2}mv^{2} = 0 \;\Rightarrow\; v = 0 \;\Rightarrow\; p = mv = 0 \]Answer: No -- \(\displaystyle PE=mgh\) is measured from the ground and is never negative, so zero mechanical energy forces \(\displaystyle KE=0\), hence \(\displaystyle v=0\) and \(\displaystyle p=0\).

  6. Exercise 15

    The power of a motor pump is 2\displaystyle 2 kW. How much water per minute the pump can raise to a height of 10\displaystyle 10 m? (Given g = 10\displaystyle 10 m s−2\displaystyle s^{-2})

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    NCERT’s answer
    m W mgh P t t Δ Δ ⇒ ×$\displaystyle 10$× $\displaystyle 10$ = = = 2000W = = $\displaystyle 12000$ or $\displaystyle 1200$ kg m
    \[P = \frac{W}{t} = \frac{mgh}{t} \] \[m = \frac{Pt}{gh} \] \[m = \frac{2000\,\text{W} \times 60\,\text{s}}{10\,\text{m s}^{-2} \times 10\,\text{m}} \] \[m = 1200\,\text{kg} \]Answer: \(\displaystyle 1200\,\text{kg}\) (\(\displaystyle 1200\) litres) of water raised each minute.

  7. Exercise 16

    The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4\displaystyle 0.4 m height on the surface of the earth. How high he can jump on the planet A?

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    NCERT’s answer
    Since, weight of the person on planet A is half that on the earth, acceleration due to gravity there, will be $\displaystyle 1$/$\displaystyle 2$ that on the earth. Hence he can jump double the height with the same muscular force. or The potential energy of the person will remain the same on the earth and on planet A. Thus, m \(\displaystyle g_{1}\)\(\displaystyle h_{1}\) = m \(\displaystyle g_{2}\)\(\displaystyle h_{2}\) if \(\displaystyle g_{1}\) = g then \(\displaystyle g_{2}\) g = , \(\displaystyle h_{1}\) = $\displaystyle 0.4$ Then \(\displaystyle h_{2}\) = g h g g g ×$\displaystyle 0.4$ = or \(\displaystyle h_{2}\) = $\displaystyle 0.4$ × $\displaystyle 2$ = 0.8m
    Weight on planet A is half that on Earth, so for the same mass, \(\displaystyle g_A = \tfrac{1}{2} g_E\). The same muscular effort gives the same take-off kinetic energy on either surface, which converts entirely to gravitational potential energy at the top of the jump: \[\frac{1}{2}mv^2 = mg_E h_E = mg_A h_A \] \[h_A = h_E \times \frac{g_E}{g_A} = 0.4\ \text{m} \times 2 = 0.8\ \text{m} \]Answer: \(\displaystyle h_A = 0.8\ \text{m}\)

  8. Exercise 17

    The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

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    NCERT’s answer
    \(\displaystyle v^{2}\) - \(\displaystyle u^{2}\) = $\displaystyle 2$ a.s This gives s= v u a - F= m a we can write work done (W) by this force F as W= ma v u a         - m v − = $\displaystyle 2$ m \(\displaystyle u^{2}\) = (K.E)f - (K.E)i
    Let the body of mass \(\displaystyle m\) move a distance \(\displaystyle s\) under the constant force \(\displaystyle F\), its speed rising from \(\displaystyle u\) to \(\displaystyle v\).\[F = ma \;\Rightarrow\; a = \frac{F}{m} \] \[v^{2} = u^{2} + 2as \] \[v^{2} - u^{2} = \frac{2Fs}{m} \] \[Fs = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} \] \[W = Fs = KE_f - KE_i \]Answer: \(\displaystyle W = Fs = \Delta KE\) -- the work done equals the gain in kinetic energy.

  9. Exercise 18

    Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.

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    NCERT’s answer
    Yes, it is possible, if an object is moving in a circular path. Because force is always acting perpendicular to the direction of displacement.
    \[W = Fs\cos\theta \] \[\theta = 90^\circ \;\Rightarrow\; \cos 90^\circ = 0 \;\Rightarrow\; W = 0 \]In uniform circular motion the centripetal force stays perpendicular to the velocity at every instant -- it changes direction, producing acceleration, but does no work.NCERT_Solution_Class9_Science_Exemplar_Ch11_Q18Answer: Yes -- e.g. a stone whirled in a circle: centripetal force \(\displaystyle \perp\) velocity gives \(\displaystyle W=0\) though acceleration \(\displaystyle \neq 0\).

  10. Exercise 19

    A ball is dropped from a height of 10\displaystyle 10 m. If the energy of the ball reduces by 40\displaystyle 40% after striking the ground, how much high can the ball bounce back? (g = 10\displaystyle 10 m s−2\displaystyle s^{-2})

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    NCERT’s answer
    m g h = m ×$\displaystyle 10$ ×$\displaystyle 10$ = 100m J. Energy is reduced by $\displaystyle 40$% then the remaining energy is 60m J. Therefore, $\displaystyle 60$ m = m ×$\displaystyle 10$ × h′ or h′ = $\displaystyle 6$ m
    \[E_i = mgh = m(10)(10) = 100m\,\text{J} \] \[E_f = (1-0.40)E_i = 0.6 \times 100m = 60m\,\text{J} \] \[mgh' = E_f \;\Rightarrow\; h' = \frac{60m}{mg} = \frac{60}{10} \] \[h' = 6\,\text{m} \]Answer: The ball rebounds to a height of \(\displaystyle 6\,\text{m}\).

  11. Exercise 20

    If an electric iron of 1200\displaystyle 1200 W is used for 30\displaystyle 30 minutes everyday, find electric energy consumed in the month of April.

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    NCERT’s answer
    P = $\displaystyle 1200$ $\displaystyle 1.2$ kW $\displaystyle 1000$ = t = 0.5h = E = Power × time × days = $\displaystyle 1.2$×.5×$\displaystyle 30$ = $\displaystyle 18$ kW h Long Answer Questions
    Energy used equals power times time; scale the daily energy by the days in April.\[P = 1200\ \text{W}, \qquad t = 30\ \text{min} = 0.5\ \text{h} \] \[E_{\text{day}} = Pt = 1200\ \text{W} \times 0.5\ \text{h} = 600\ \text{Wh} = 0.6\ \text{kWh} \] \[E_{\text{April}} = E_{\text{day}} \times 30\ \text{days} = 0.6\ \text{kWh} \times 30 = 18\ \text{kWh} \]Answer: \(\displaystyle 18\ \text{kWh}\).