Question

A(n) 14 g bullet is fired into a(n) 121 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 8.3 m before coming to rest. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of friction between the surface and the block is 0.7, find the speed of the bullet before impact. Answer in units of m/s

Answer 1

Answer:

33.14 m/s

Explanation:

The mass of the block is 121g or .121 kg. As the bullet is lodged in the block the total mass is 121+14 = 135 g or 0.135 kg.

The frictional force that makes the block come to a stop is normal force* coefficient of friction = 0.135 * 9.8 * 0.7 = 0.9261 N

As the block comes to rest after sliding for 8.3 meters the energy it was given by the bullet is

0.135 * 9.8 * 0.7 * 8.3

= 7.69 Nm

Now this energy is provided the bullet. So the energy in the bullet was equal to

1/2 * mv² = 0.5 * 14 * v².

0.5 * 0.014 * v^2 = 0.135 * 9.8 * 0.7 * 8.3 = 7.69

=> 0.007 * v² = 7.69

=> v² = 7.69 / 0.007

=> v² = 1098.57

=> v = √1098.57

=> v = 33.14 m/s