Question

A block of weight 100N is pushed by a force F on a horizontal 1 m/s2, when force is
doubled its acceleration becomes 10 m/s2. The coefficient of friction is _(g = 10ms-2)
(a) 0.2
(b) 0.4
(C) 0.6
(d) 0.8​

Answer 1

Answer:

Given that,

Weight, W=100N

Acceleration, a=1m/s

2

Acceleration when force is double, a=10m/s

2

Now, Let the External Force be F and friction force be f.

Then,

F−f

s

=ma

F−μN=ma

F−μ100=10×1

F−μ100=10....(I)

We know, When Force is doubled,

Then,

2F−f

s

=ma

2F−μN=ma

2F−μ100=10×10

2F−μ100=100....(II)

Now, from equation (I) and (II)

F=90N

Now, put the value of F in equation (I)

90−μ100=10

−μ100=10−90

μ=

100

80

μ=0.8

Hence, the coefficient of restitution is 0.8