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 2.7m along the table top and comes to a stop. The force of friction between the block and the table is

Options

A. 0.052 N
B. 3.63 N
C. 2.50 N
D. 1.04 N

Answer

A

Explanation

Velocity of block just after collision will be
`v = (5 xx 10^(-3) xx 150)/(2+5 xx 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 xx 2.7 = 1/2 xx 2.005 xx (0.374)^(2)` or `F=0.052N`