Question

A rifle bullet, of mass 18.00 g, is fired to the right at a speed of 240.00 m / s to hit a sack of sand and penetrate it 35.00 cm. Determine the magnitude of the friction force (assumed constant) acting on the bullet.

a) –6.17 N

b) 5924.57 N​​​​​​​

c) 1481.14 N​​​​​​​

d) –12.34 N​​​​​​​

e) None of the above.

Answer 1

To determine the constant friction force acting on a bullet as it penetrates a sandbag, we can use the work-energy principle.

Given:

• Mass of the bullet, $m = 18.00 \, \text{g} = 0.018 \, \text{kg}$

• Initial velocity of the bullet, $v_i = 240.00 \, \text{m/s}$

• Final velocity of the bullet, $v_f = 0 \, \text{m/s}$ (since it comes to rest)

• Penetration distance, $d = 35.00 \, \text{cm} = 0.35 \, \text{m}$

Solution:

1. Calculate the initial kinetic energy (KE) of the bullet:

   $$\text{KE} = \frac{1}{2} m v_i^2$$$$\text{KE} = \frac{1}{2} \times 0.018 \, \text{kg} \times (240.00 \, \text{m/s})^2$$$$\text{KE} = 0.009 \times 57600$$$$\text{KE} = 518.4 \, \text{J}$$

2. Apply the work-energy principle:

   The work done by the friction force ($F_{\text{friction}}$) brings the bullet to rest. This work is equal to the initial kinetic energy of the bullet.

   $$\text{Work} = F_{\text{friction}} \times d$$

   Since the work done by friction is equal to the initial kinetic energy:

   $$F_{\text{friction}} \times d = \text{KE}$$

   Solving for $F_{\text{friction}}$:

   $$F_{\text{friction}} = \frac{\text{KE}}{d}$$$$F_{\text{friction}} = \frac{518.4 \, \text{J}}{0.35 \, \text{m}}$$$$F_{\text{friction}} \approx 1481.14 \, \text{N}$$

Conclusion:

The magnitude of the constant friction force acting on the bullet as it penetrates the sandbag is approximately 1481.14 N.

Answer 2