A72.3-kg person, running horizontally with a velocity of +2.99 m/s, jumps onto a 13.9-kg sled that is initially at rest. (a) ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) the sled and person coast 30.0 m on level snow before coming to rest. what is the coefficient of kinetic friction between the sled and the snow?

(a) 2.51 m/s

(b) 0.011

Explanation:

mass of person, M = 72.3 kg

initial velocity of person, U = 2.99 m/s

mass of sled, m = 13.9 kg

initial velocity of sled, U = 0

(a) Let the velocity of person and the sled is V

By using the conservation of momentum

M x U + m x u = (M + m) x V

72.3 x 2.99 + 13.9 x 0 = (72.3 + 13.9) V

216.177 = 86.2 V

V = 2.51 m/s

(b) initial velocity , V = 2.51 m/ s

Final velocity, V' = 0

distance, s = 30 m

Let μ be the coefficient of friction.

Use third equation of motion

0 = 2.51 x 2.51 - 2 x μ x 9.8 x 30

μ = 0.011

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.