SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Work, Energy, and Simple Machines

28 questions · 20 still being checked

Revise, Reflect, Refine 7.12–7.13 (part 19 of 21)

  1. Exercise 7.12

    The gravitational attraction on the surface of the Moon (lunar surface) is about 1\displaystyle 1 th An astronaut can throw a ball up to a height of 8\displaystyle 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
    NCERT’s answer
    $\displaystyle 48$ m
    The ball rises $\displaystyle 48$ m on the Moon — six times as high.
    NCERT_Solution_Class9_Science_Ch7_RRR_Q7-12
    At the highest point all the kinetic energy has become potential energy:
    \(\displaystyle \frac{1}{2}mv^{2} = mgh \), so \(\displaystyle h = \dfrac{v^{2}}{2g} \)
    The astronaut throws with the same upward velocity \(\displaystyle v \), so \(\displaystyle h \) is inversely proportional to \(\displaystyle g \):
    \(\displaystyle \dfrac{h_{\text{Moon}}}{h_{\text{Earth}}} = \dfrac{g_{\text{Earth}}}{g_{\text{Moon}}} = 6 \) (since \(\displaystyle g_{\text{Moon}} = \frac{1}{6} g_{\text{Earth}} \))
    Therefore \(\displaystyle h_{\text{Moon}} = 6 \times 8\ \text{m} = 48\ \text{m} \).
    The mass of the ball cancels out, so the answer does not depend on how heavy the ball is.

  2. Exercise 7.13

    A 1000\displaystyle 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?NCERT_Question_Class9_Science_Ch7_RRR_Q7-13

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i) Between A and B the car moves at constant speed, \(\displaystyle 35\ \text{m s}^{-1} \).
    The speed-time graph is a horizontal line from A to B, so the speed does not change and the acceleration is zero.
    This is the driver's reaction time — the obstruction has been spotted but the brakes have not been applied yet.
    (ii) Kinetic energy at A = $\displaystyle 612500$ J.
    \(\displaystyle K = \frac{1}{2}mv^{2} = \frac{1}{2} \times 1000\ \text{kg} \times (35\ \text{m s}^{-1})^{2} \)
    \(\displaystyle K = 500 \times 1225 = 612500\ \text{J} \)
    (iii) Work done by the brakes between B and C = \(\displaystyle -612500\ \text{J} \).
    By the work-energy theorem, work done = change in kinetic energy \(\displaystyle = 0\ \text{J} - 612500\ \text{J} = -612500\ \text{J} \)
    It is negative because the braking force acts opposite to the car's displacement.
    (iv) The kinetic energy is converted mainly into thermal energy in the brake pads, tyres and road, with a small part leaving as sound.
    Braking is friction, and the chapter states plainly that work done against friction does not lead to any storage of energy — so none of this energy is stored as potential energy.
    The car runs on a level road, so its height never changes and \(\displaystyle U = mgh \) stays fixed at its ground-level value; there is no rise for gravitational potential energy to be gained through.
    Thermal energy is one of the forms of energy listed in the chapter, and the chapter records that mechanical energy and thermal energy can be converted into one another, and that energy can be transferred as heat.
    This is the same reason the chapter gives for a real pendulum slowing and stopping, and for the roller-coaster ball reaching lower and lower heights: the energy is carried away by friction and air resistance, not banked as potential energy.