Question

a bullet of mass m is fired into a block of mass m that is at rest. the block, with the bullet embedded, slides distance d across a horizontal surface. the coefficient of kinetic friction is μk.

Answer 1

Uk=2m*the normal force
the distance d is needed only if you were asked about work

Answer 2

The expression for the bullet's speed vbullet is $v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}$ and  the speed of a 9.0 g bullet is v=16.8 m/s

Explanation:

A  bullet of mass m is fired into a block of mass m that is at rest. the block, with the bullet embedded, slides distance d across a horizontal surface. the coefficient of kinetic friction is μk.

The initial kinetic energy of the bullet is given by

$K_i = \frac{1}{2}mv^2$

where

m is the mass of the bullet  

v is the initial speed of the bullet

The expression for the bullet's speed vbullet is

$F=\mu_k (m+M) g$

where

$\mu_k$ is the coefficient of kinetic friction

g is the acceleration of gravity

The work done by the force of friction is

$W=-Fd = -\mu_k (m+M)g d$

where d is the displacement of the block+bullet.

Because the final kinetic energy is zero (the bullet with the block comes at rest), we can write:

$W=K_f - K_i = -K_i$

And so

$-\mu_k (m+M) g d = -\frac{1}{2}mv^2$

By solving for v, the solution for the bullet speed:

$-\mu_k (m+M) g d = -\frac{1}{2}mv^2\\v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}$

The speed of a 9.0 g bullet that, when fired into a 12 kg stationary wood block causes the block to slide 5.4 cm across a wood table. Assume that k=0.20.

We have:

the mass of the bullet, m = 9.0 g = 0.009 kg

the mass of the block, M = 12 kg

the distance covered by the block+bullet, d = 5.4 cm = 0.054 m

the coefficient of friction, $\mu_k = 0.20$

the acceleration of gravity, $g = 9.8 m/s^2$

By substituting, we got

$v=\sqrt{\frac{2 (0.20) (0.009 kg+12 kg)(9.8 m/s^2)(0.054 m)}{0.009 kg}}=16.8 m/s$

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