Piggy, the title did not show the end, randomly analyze the principle for you well.

Objects sliding on the B, relative to the point on the B speed is divided into horizontal to the left of v and backward v, the combined speed √2v is along the direction of 45 ° with the B, then the trajectory of movement relative to the B is certainly a straight line.

Assuming that it is subjected to a sliding friction force f=μmg,the direction is in the same straight line as the combined relative velocity, so that the angle θ=45°, the acceleration with respect to B is also along this direction,

After t it slides to the midline of B and comes to rest with respect to B. μgt=√2v,μg=√2v/t,and the distance of motion L=? √2?d/2.?

And L=μgt?/2,Substitute L and μg so t=d/v.

μ=√2v?/gd,

Before sliding on to B, the absolute velocity of the workpiece is v and the kinetic energy is mv?/2,

After sliding on to B and coming to a relative standstill, the absolute velocity is also v, and the kinetic energy is mv?/2,

And sliding on top of B Only the friction force does work, and the kinetic energy does not change, so the friction force of B on the workpiece does 0