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Question: A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring c...

A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring constant 100 N m-1-1 as shown in the figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. The coefficient of friction between the block and the incline is:

(Take g=10 ms2\mathrm { g } = 10 \mathrm {~ms} ^ { - 2 } and assume that the pulley is frictionless)

A

0.2

B

0.3

C

0.5

D

0.6

Answer

0.3

Explanation

Solution

Here, m = 1 kg,

From figure,

N=mgcosθ\mathrm { N } = \mathrm { mg } \cos \theta

Where μ\muis the coefficient between the block and the incline.

Net force on the block down the incline,

=mgsinθμmgcosθ=mg(sinθμcosθ)= m g \sin \theta - \mu m g \cos \theta = m g ( \sin \theta - \mu \cos \theta )

Distance moved, x =10 cm = 10 × 10-2m

In equilibrium.

Work done = potential energy of stretched spring

sin45μcos45=12\sin 45 ^ { \circ } - \mu \cos 45 ^ { \circ } = \frac { 1 } { 2 }

12μ2=12\frac { 1 } { \sqrt { 2 } } - \frac { \mu } { \sqrt { 2 } } = \frac { 1 } { 2 }

μ=0.3\mu = 0.3