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Difficulty: MediumWork, Energy and Power

A bullet of mass 20 g20\text{ g} is moving horizontally at a speed of 400 m s−1400\text{ m s}^{-1}. It strikes a stationary block of wood and emerges from the opposite side with a speed of 100 m s−1100\text{ m s}^{-1}. What is the magnitude of the work done by the bullet against the resistance of the block, in Joules?

Answer: 1500 J

Answer

The magnitude of the work done by the bullet in penetrating the block is 1500 J1500\text{ J}.
According to the Work-Energy Theorem, the net work done on an object equals the change in its kinetic energy. The reduction in kinetic energy as the bullet slows from 400 m s−1400\text{ m s}^{-1} to 100 m s−1100\text{ m s}^{-1} equals the work done against the resistive force of the wooden block.

Step-by-Step Solution

1
Convert the mass of the bullet into standard SI units (kilograms)
m=201000=0.02 kgm = \frac{20}{1000} = 0.02\text{ kg}
The standard SI unit for mass in energy equations is the kilogram.
2
Determine the initial kinetic energy of the bullet before entering the block
Eki=12×0.02 kg×(400 m s−1)2=1600 JE_{ki} = \frac{1}{2} \times 0.02\text{ kg} \times (400\text{ m s}^{-1})^2 = 1600\text{ J}
Kinetic energy is defined as Ek=12mv2E_k = \frac{1}{2} m v^2.
3
Determine the final kinetic energy of the bullet as it emerges from the block
Ekf=12×0.02 kg×(100 m s−1)2=100 JE_{kf} = \frac{1}{2} \times 0.02\text{ kg} \times (100\text{ m s}^{-1})^2 = 100\text{ J}
The bullet loses speed upon passing through the block, reducing its kinetic energy.
4
Apply the Work-Energy Theorem to find the work done against resistive forces
W=ΔEk=1600 J−100 J=1500 JW = \Delta E_k = 1600\text{ J} - 100\text{ J} = 1500\text{ J}
The net work done on the bullet equals its change in kinetic energy.

Key Concept

Work-Energy Theorem
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Topics

Work, Energy and Power

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