Answer:
Explanation:
In order for the boy to start himself and the crate moving from rest, the boy would have to apply a force to the ground, which applies a reverse force that pushes the boy and the crate.
This force is opposite of static friction, in the positive direction
f = μs * m1g
μs can be 0 - 0.800, but since the boy sometimes slip, we can assume that u1_s = 0.800 where he is applying a force just at the maximum point of static friction.
This force must overcome the kinetic friction of the box in order to keep it moving (assuming it is already in motion). This frictional force actions towards the negative direction.
F = -μk * m2g
If Fnet = μs*m1g - μk*m2 = 0, assuming the crate is already in motion, this is just enough force to have 0 acceleration, thus keep the crate in motion. We can use this equation to solve for m2.
m2 = μs*m1 /μk
= 0.90 * 48.0 / 0.200
= 216kg.