Question

A cord is used to vertically lower an initially stationary block of mass M = 20 kg at a constant downward acceleration of g/8. When the block has fallen a distance d = 3.5 m

A. find the work done by the cord's force on the block.

B. Find the kinetic energy of the block

C. Find the speed of the block

Answer 1

Using f = M*a , consider a FBD of the block M*a = - M * g / 8 = T - M*g

The tension in the cord is T

T = M*g - M*g / 8

T = 7 M g / 8

Work done by the T is - 7 M * g * d / 8 = -7 * 20 * 9.8 * 3.5 / 8 = -600.25 =600.25 N in opp direction

Work done by the weight of the block is = M * g* d

the net is equal to kinetic energy , or M * g * d / 8 = 20* 9.8 * 3.5 / 8 = 85.75 kg m^2/ s^2 = 85.75 J

This is related to the speed of the block , V , as

0.5* M * v^2 = M * g* d / 8

V = 2 * sqrt ( g * d )

v = 2 * sqrt ( 9.8 * 3.5 )

V = 11.7 m/s