Question

a bullet of mass 0.04kg mpoving with a speed of 90meters per second square enters a heavy wooden block and is stopped after a distance of 60 cm. what is the average resistive force exerted by the block on the bullet??

Answer 1

A bullet of mass 0.04kg moving with a speed of 90meters per second square enters a heavy wooden block and is stopped after a distance of 60 cm. what is the average resistive force exerted by the block on the bullet?

**Solution:**

$$V = 90 \frac{m}{sec}, m = 0.04 \, kg, L = 0.6 \, m$$

Kinetic energy of the bullet at the beginning and end of the movement:

$$E_1 = \frac{mV^2}{2}$$

$E_2 = \frac{mV^2}{2} = 0$ (bullet stopped, velocity is zero, then the kinetic energy is also zero)

The theorem on the change of kinetic energy: change of kinetic energy of the system is the work of all internal and external forces acting on the body system.

$$E_2 - E_1 = A \, (1)$$

The only force that acts on the bullet - the resistance force of wooden block:

$$A = \overrightarrow{F_{res}} \cdot \vec{s} = - F_{res} L \, (2)$$

(minus sign because direction of the force against the motion bullet)

$$(2) \, in \, (1): 0 - \frac{mV^2}{2} = - F_{res} L$$ $$\frac{mV^2}{2} = F_{res} L$$ $$F_{res} = \frac{mV^2}{2L} = \frac{0.04 \, kg \cdot (90 \, \frac{m}{sec})^2}{2 \cdot 0.6 \, m} = 270 \, N$$

Answer: $F = 270 \, N$