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Work
Question

A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring constant 100Nm1 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=10ms2 and assume that the pulley is frictionless)

Moderate
Solution

Here, m=1kg,θ=45,k=100Nm1

From figure, N=mgcosθ

f=μN=μmgcosθ

where μ is the coefficient of friction between the block and the incline.

Net force on the block down the incline,

=mgsinθf

=mgsinθμmgcosθ=mg(sinθμcosθ)

Distance moved, x=10cm=10×102m

Work done = Potential energy of stretched spring 

mg(sinθμcosθ)x=12kx2

2mg(sinθμcosθ)=kx

2×1×10×sin45μcos45=100×10×102

sin45μcos45=1212μ2=12

1μ=22=12  μ=112=212

 μ=0.3

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