First slide

Kinetic friction

Question

A 2 kg block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of $150 {ms}^{- 1}$ is shot into the block and sticks to it. The block then slides 2.7 m along the table top and comes to a stop. The force of friction between the block and the table is

Moderate

A

0.052 N
B

3.63 N
C

2.50 N
D

1.04 N

Solution

Velocity of block just after collision,
$v = \frac{5 \times 10^{- 3} \times 150}{(2 + 5 \times 10^{- 3})}$
(from conservation of linear momentum)
$= 0.374 {ms}^{- 1}$
Let F be the force of friction, then work done against friction = initial kinetic energy
or $F \times 2.7 = \frac{1}{2} \times 2.005 \times (0.374)^{2} \Rightarrow F = 0.052 N$

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