Q.A bullet of mass $0.04\ \text{kg}$ moving with a speed of $90\ \text{m s}^{-1}$ enters a heavy wooden block and is stopped after a distance of $60\ \text{cm}$. What is the average resistive force exerted by the block on the bullet?

Concept: Work-Energy Theorem — the work done by the net force equals the change in kinetic energy.

The bullet’s initial kinetic energy is

$K_i = \frac{1}{2} m v^2 = \frac{1}{2} (0.04)(90)^2 = 162\ \text{J}$.

It comes to rest, so $K_f = 0$. The work done by the resistive force $F$ (opposing motion) over a displacement $s = 0.60\ \text{m}$ is

$W = -F s$ (negative because force opposes displacement). …

The key here is to avoid getting tangled in kinematics (finding acceleration, then force). The Work-Energy Theorem gives a direct path: the net work done on an object equals its change in kinetic energy. Since the bullet starts moving and ends at rest, all its initial kinetic energy is dissipated by the resistive force of the wood.

The only force doing work on the bullet (once it enters the block) is the resistive force $F$, which acts opposite to the direction of motion. So the work done by this force is negative: $W = -F \times d$, where $d$ is the penetration distance.

Let’s walk through it step by step.

1. Identify the known quantities

   Mass of bullet: $m = 0.04\ \text{kg}$

   Initial speed: $u = 90\ \text{m/s}$

   Final speed: $v = 0\ \text{m/s}$

   Penetration distance: $s = 60\ \text{cm} = 0.6\ \text{m}$

   Resistive force: $F$ (to be found, assumed constant on average)

2. Write the initial kinetic energy

Compute stepwise: $90^2 = 8100$, then $\frac{1}{2} \times 0.04 = 0.02$, so

3. Final kinetic energy

   Since the bullet stops, $K_f = 0\ \text{J}$.

4. Change in kinetic energy

5. Work done by the resistive force The force $F$ acts opposite to displacement, so work done by it is:

6. Apply the Work-Energy Theorem

$-F \times 0.6 = -162$ …

Concept: Equations of Motion (kinematics) + Newton's Second Law

Step 1: Write down the known quantities

$m = 0.04\ \text{kg}$, $u = 90\ \text{m s}^{-1}$, $v = 0$ (bullet stops), $s = 60\ \text{cm} = 0.6\ \text{m}$.

Step 2: Find the deceleration using $v^2 = u^2 + 2as$

The negative sign shows the acceleration opposes the bullet's motion (it is a deceleration).

Step 3: Apply Newton's Second Law to get the force

Step 4: State the direction …