A 1kg block situated on a rough incline is connected to a spring of spring constant `100Nm^(-1)` as shown in figure,. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless.

Options

A. 0.2
B. 0.3
C. 0.5
D. 0.6

Answer

B

Explanation

Here, m=1 kg, `theta=45^@, k=100 N m^(-1)`
From figure, `N=mgcostheta`
`f=muN=mumgcos theta`
where `mu` is the coefficient of friction between the block and the incline.
Net force on the block down the incline.
`=mgsin theta-f`
`=mgsintheta-mumgcostheta=mg(sintheta-mucostheta)`
Distance moved, x=10 cm =`10 xx 10^(-2)` m
In equilibrium,
Work done = Potential energy of stretched spring
`mg (sin theta -mu cos theta)x =1/2 kx^2`
`2mg(sin theta-mucostheta) =kx`
`2xx1xx10xx(sin 45^@ - mucos45^@) = 100 xx 10 xx 10^(-2)`
`sin 45^@- mu cos 45^@=1/2`
`1/sqrt2-mu/sqrt2=1/2`
`1-mu=sqrt2/2=1/sqrt2implies mu=1-1/sqrt2=(sqrt2-1)/sqrt2`
`mu=0.3`