A 5.00 gg bullet is fired horizontally into a 1.20 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 mm along the surface before stopping.

V= 295.2m/s

Explanation:

Given that mass of bullet = 5.0g = 0.005kg

Mass of wooden block = 1.20kg

The coefficient of kinetic friction between block and surface = 0.20

The force of friction of the block

Fk = (μ) mg

Fk = 0.2×1.2×9.8m/s

F = 2.352N

= Force × distance

The work done = 2.352× 0.390

= 0.91728J

The initial velocity of the block can be calculated

1/2mv^2

0.5×1.2×v^2 = 0.917J

0.6V^2 = 0.917J

V^2 = 0.917J/0.6

V^2 = 1.52

V = √1.52

V = 1.23m/s

Now we shall use conservation of momentum to find the velocity of the bullet

0.005kg×v = 1.2kg × 1.23m/s

0.005v = 1.476

V= 1.476/0.005

V= 295.2m/s

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.