JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Work, Power & Energy

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A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force

F

on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (

g = 10 ms^{- 2}

) ______.

Answer: 150

Solution

The ball leaves with just enough kinetic energy to rise to

20

m, so by the work-energy theorem the launching force does work equal to that gain in energy:

F d = m g h

. With

m = 0.15

kg,

h = 20

d = 0.2

m and

g = 10

m s

^{- 2}

F = \frac{m g h}{d} = \frac{0.15 \times 10 \times 20}{0.2} = 150

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Integer answer

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