Question

In the system shown below, the block M (mass 5.65 kg) is pushed so that it is initially moving to the left with a speed of vo 1.75 m/s The hanging weight m hass a mass of 2.86 kg and the coefficients of friction are μ,-0.411 and μ,-0.304. The string and the pulley have negligible mass. How fast will M be traveling when m has fallen through a height h = 0.47 meters?

Answer 1

Given

From work energy theorem

initial energy

Ei = m*g*h + (1/2)*M*vo^2 + (1/2)*m*v0^2

work done = -uk*M*g*L

final energy Ef = (1/2)*m*v^2 + (1/2)*M*v^2

Wf = Ef - Ei

-0.304*5.65*9.8*0.47 = (0.5*(2.86+5.64)*v^2) - (2.86*9.8*0.47) - (0.5*(5.65+2.86)*1.75^2)

v = 2.07 m/s