Question

A bullet of mass 20 g, moving at 50 m/s penetrates 20 cm into a wooden block. What is the magnitude of the force exerted on the wooden block?

• A. 625 N
• B. 225 N
• C. 125 N
• D. 725 N

Hint:

To calculate the force exerted by a bullet on an object, use the work-energy principle. The force can be determined by dividing the kinetic energy by the penetration distance.

Answer 1

The Correct Option is C

Step 1: Using the work-energy principle.
The work done by the force exerted on the block is equal to the change in the kinetic energy of the bullet. The bullet is initially moving at 50 m/s, and after penetrating 20 cm into the block, it comes to rest. Therefore, the initial kinetic energy of the bullet is converted into work done in stopping the bullet. The kinetic energy of the bullet is given by: \[ K.E. = \frac{1}{2} m v^2 \] where:
\( m = 20 \, {g} = 0.02 \, {kg} \) (mass of the bullet),
\( v = 50 \, {m/s} \) (velocity of the bullet).
Substitute the values: \[ K.E. = \frac{1}{2} \times 0.02 \, {kg} \times (50 \, {m/s})^2 = \frac{1}{2} \times 0.02 \times 2500 = 25 \, {J} \] Step 2: Calculating the force.
The work done by the force in bringing the bullet to rest is equal to the force \( F \) multiplied by the distance over which the bullet penetrates the block, i.e., 20 cm = 0.2 m. The work done is: \[ W = F \times d \] where:
\( d = 0.2 \, {m} \) (distance the bullet penetrates the block),
\( W = 25 \, {J} \) (work done).
Since work done is equal to kinetic energy, we have: \[ F \times 0.2 = 25 \] Solve for \( F \): \[ F = \frac{25}{0.2} = 125 \, {N} \] Step 3: Conclusion.
Thus, the magnitude of the force exerted on the wooden block is \( 125 \, {N} \).