Question

A block of mass 200g is sitting on top of another block of three times that mass which is on a horizontal frictionless surface and is attached to a horizontal spring. The coefficient of static friction between the blocks is 0.2. The lower block is pulled until the attached spring is stretched by 5.0cm and is then released from rest. Find the maximum value of the spring constant for which the upper block does not slip on the lower block.

Answer 1

for the upper block to slip on the lower block, the force due to spring is greater than the maximum static friction force.

maximum static friction force=friction coefficient*normal force

=friction coefficient*weight of the upper block=0.2*0.2*9.8=0.392 N

let acceleration of the combined block system is a m/s^2. (just before the upper block slips)

then writing force equation for forces on the upper block:

0.392=0.2*a

==>a=0.392/0.2=1.96 m/s^2

writing force balance equation for the lower block:

spring force-friction force=mass*acceleration

==>spring constant*0.05-0.392=0.2*a=0.2*1.96

==>spring constant=(0.392+0.2*1.96)/0.05=15.68 N/m

hence maximum value of spring constant is 15.68 N/m.