SolveItClass 9 · NCERT

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

28 questions · 20 still being checked

Revise, Reflect, Refine 7.8–7.9 (part 13 of 21)

  1. Exercise 7.8

    A man of mass 60\displaystyle 60 kg rides a scooter of mass 100\displaystyle 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40\displaystyle 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
    NCERT’s answer
    $\displaystyle 4$:$\displaystyle 5$
    Ratio of fuel used, day $\displaystyle 1$ : day $\displaystyle 2$ = $\displaystyle 4$ : $\displaystyle 5$.
    NCERT_Solution_Class9_Science_Ch7_RRR_Q7-8
    Day $\displaystyle 1$ — mass set in motion = scooter + man \(\displaystyle = 100\ \text{kg} + 60\ \text{kg} = 160\ \text{kg} \)
    Energy supplied \(\displaystyle E_{1} = \frac{1}{2}m_{1}v^{2} = \frac{1}{2} \times 160\ \text{kg} \times v^{2} = 80v^{2} \)
    Day $\displaystyle 2$ — mass set in motion \(\displaystyle = 100\ \text{kg} + 60\ \text{kg} + 40\ \text{kg} = 200\ \text{kg} \)
    Energy supplied \(\displaystyle E_{2} = \frac{1}{2} \times 200\ \text{kg} \times v^{2} = 100v^{2} \)
    All this energy comes from the fuel and nothing is lost to air resistance or friction, so the fuel used is in the same ratio as the energy:
    \(\displaystyle \dfrac{E_{1}}{E_{2}} = \dfrac{80v^{2}}{100v^{2}} = \dfrac{4}{5} \)
    The final speed is the same on both days, so the extra fuel is needed only because of the extra $\displaystyle 40$ kg that must be given kinetic energy.

  2. Exercise 7.9

    On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The child sits twice as far from the fulcrum as the adult — for example, child at $\displaystyle 2$ m and adult at $\displaystyle 1$ m.
    NCERT_Solution_Class9_Science_Ch7_RRR_Q7-9
    A balanced seesaw is a lever obeying \(\displaystyle \text{effort} \times \text{effort arm} = \text{load} \times \text{load arm} \) (Eq. $\displaystyle 7.15$).
    Let the child's weight be \(\displaystyle W \) at distance \(\displaystyle d_{c} \), and the adult's weight \(\displaystyle 2W \) at distance \(\displaystyle d_{a} \):
    \(\displaystyle W \times d_{c} = 2W \times d_{a} \)
    \(\displaystyle d_{c} = 2 d_{a} \)
    What your figure must show:
    A horizontal plank resting on a triangular fulcrum at its middle, drawn level to show it is balanced.
    The adult seated on the right, with the distance from the fulcrum marked \(\displaystyle d = 1\ \text{m} \), and a downward arrow labelled \(\displaystyle 2W \).
    The child seated on the left, with the distance marked \(\displaystyle 2d = 2\ \text{m} \), and a downward arrow labelled \(\displaystyle W \).
    Label the three parts — fulcrum, load arm and effort arm — and write \(\displaystyle W \times 2\ \text{m} = 2W \times 1\ \text{m} \) beneath the drawing.
    Any pair of distances in the ratio $\displaystyle 2$ : $\displaystyle 1$ (child : adult) is correct; the sliding seats are what let them find it.