A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring constant 100Nm−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=10ms−2 and assume that the pulley is frictionless)

Here, m=1kg,θ=45∘,k=100Nm−1From figure, N=mgcos⁡θf=μN=μmgcos⁡θwhere μ is the coefficient of friction between the block and the incline.Net force on the block down the incline,=mgsin⁡θ−f=mgsin⁡θ−μmgcos⁡θ=mg(sin⁡θ−μcos⁡θ)Distance moved, x=10cm=10×10−2mWork done = Potential energy of stretched spring mg(sin⁡θ−μcos⁡θ)x=12kx22mg(sin⁡θ−μcos⁡θ)=kx2×1×10×sin⁡45∘−μcos⁡45∘=100×10×10−2sin⁡45∘−μcos⁡45∘=12⇒12−μ2=121−μ=22=12 ⇒ μ=1−12=2−12∴ μ=0.3