Question

A block of wood with a mass of = 2 kg rests on a tabletop with a coefficient of friction between the block and the table. A = 10 g bullet is shot horizontally with the velocity 1 = 800 /. The bullet hits the block, makes an end-to-end hole in it and exits the block with a velocity 2 = 200 / from its other side. The collision (impact) is very short, almost instantaneous. As a result of such an impact, the block slides the distance = 1 from its initial position on the tabletop and stops. _________________________________________________________ a) What part of the initial energy of the bullet was transformed to heat during the impact?
b) Find the coefficient of friction .
c) What was the velocity of the block immediately after the impact and at half distance (/2) before it stops?

Answer 1

The force of friction between the block and the table is F = 0.052 N

Explanation:

Given data:

Mass of block "m" = 2 Kg

Mass of bullet "m" = 5 g

Distance covered by block =  2.7 m

Solution:

Velocity of block just after collision will be

v = 5×10^−3×150 / 2+5×10^−3

v = 750 x 10^-3 / 7 x 10^-3

v = 0.374 m /s

Let, F be the force of friction. Then, work done against friction = initial kinetic energy.

F × 2.7 = 1 / 2 × 2.005×(0.374)^2

F = 0.052 N

Hence the force of friction between the block and the table is F = 0.052 N