A bullet of mass 2.5 g moving with a velocity of $500 m s^{- 1}$ , enters a wooden block and comes out of it with a velocity of $100 m s^{- 1}$ . Find the work done by the bullet while passing through the wooden block.

100 J

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300 J

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500 J

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800 J

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Solution

The correct option is B 300 J
We know,

Work done=Change in Kinetic Energy

$Δ K E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

$Δ K E = W = \frac{1}{2} \times \frac{2.5}{1000} \times 100^{2} (5^{2} - 1)$

$W = 0.5 \times 25 \times 24 = 300 J$

Option $B$ is the correct answer

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A bullet of mass 2.5 g moving with a velocity of 500ms−1, enters a wooden block and comes out of it with a velocity of 100ms−1. Find the work done by the bullet while passing through the wooden block.

The correct option is B 300 JWe know,Work done=Change in Kinetic EnergyΔKE=12m(v2f−v2i)ΔKE=W=12×2.51000×1002(52−1)W=0.5×25×24=300JOption B is the correct answer

A bullet of mass 2.5 g moving with a velocity of $500 m s^{- 1}$ , enters a wooden block and comes out of it with a velocity of $100 m s^{- 1}$ . Find the work done by the bullet while passing through the wooden block.

The correct option is B 300 J
We know,

Work done=Change in Kinetic Energy

$Δ K E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

$Δ K E = W = \frac{1}{2} \times \frac{2.5}{1000} \times 100^{2} (5^{2} - 1)$

$W = 0.5 \times 25 \times 24 = 300 J$

Option $B$ is the correct answer