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 isA. 0.052 NB. 3.63 NC. 2.50 ND. 1.04 N

A 2 kg block of wood rests on a long table top. A 5 g bullet moving ho

1.	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 isA. 0.052 NB. 3.63 NC. 2.50 ND. 1.04 N
Answer» Correct Answer - A 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`

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