Question

A block having mass 500 g slides on a rough horizontal table, if the friction coefficient between block
and table is 0.2 and initial speed of the block is 60 cm/s. Then calculate:
Work done by frictional force in bringing the block to rest.
How far does the block move before coming to rest. (g = 10 m/s)​

Answer 1

Answer:

Work w =0.9 J

S= 0.09 m

Explanation:

Given that

m = 500 g⇒m=0.5 Kg

Friction coefficient μ=0.2

Initial velocity = 60 cm/s⇒ u=0.6 m/s

When block will move, then friction force will act in opposite direction of motion and will try to stopped the motion of the block.So the retardation offered by friction will be gμ.

Lets take distance travel by block before come in to rest is S.

We know that

$v^2 = u^2+2 a S$

At final state v =0

So

$v^2 = u^2+2 a s$

0=0.6 x 0.6 -2 x 0.2 x 10 x S

s= 0.09 m

The work done by friction force

W= Friction force x distance             (fr =  μmg)

W= 0.2 x 0.5 x 10 x 0.09

Work W =0.9 J