Question

A 40kg slab rests on a fraction less floor . A 20kg block rests on the top of the slab. The coefficient of static friction between the block and the slab is 0.6 and the coefficient of kinetic friction is 0.4 . The 20kg block is acted upon by a horizontal force of 200N. Then resulting acceleration of slab is (g=10m/s2)

Answer 1



Secondary SchoolPhysics5+3 pts

A 40kg slab rests on a frictionless floor . A 10kg block rests on top of slab . coefficient of kintic friction between the block and slab is 0.4 . A horizontal force of 100 N applied on 10 kg block .Find resulting acceleration of slab

Report by Uday22271 29.01.2017

Answers





Duragpalsingh Maths AryaBhatta

Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N 
Static frictional force = 98.1 * 0.6 N is less than 100 N applied 
=> 10 kg blck will slide on 40 kg slab and net force on it 
= 100 N - kinetic friction 
= 100 - 98.1 * 0.4 = 61 N 
=> 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s 

Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N 
=> 40 kg slab will move with 
39/40 m/s