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 (integer only)

$[g = 10 {ms}^{- 2}]$

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Solution

From work energy theorem,

$W = Δ K E$

$F . x = \frac{1}{2} m v^{2}$

Here, $v^{2} = 2 g h$

$F \times \frac{2}{10} = \frac{1}{2} \times \frac{15}{100} \times 2 \times 10 \times 20$

$∴ F = 150 N$

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A ball of mass 0.2kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes up to 2m height further, find the magnitude of force. (Consider g = 10m/s²).

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From work energy theorem, W=ΔKE F.x=12mv2 Here, v2=2gh F×210=12×15100×2×10×20 ∴F=150 N

From work energy theorem,

$W = Δ K E$

$F . x = \frac{1}{2} m v^{2}$

Here, $v^{2} = 2 g h$

$F \times \frac{2}{10} = \frac{1}{2} \times \frac{15}{100} \times 2 \times 10 \times 20$

$∴ F = 150 N$