The normal reaction force of block on incline R = mg cosθ
Let the coefficient of friction be M
∴ F = MR
= μ mg cosθ
∴ Net force on block = mg sinθ – F
= mg (sinθ – μ cosθ)
we know that the block moves by distance x = 10cm = 0.1 m.
The work done by net force in moving block 0.1m is
= Energy stored in the spring.
∴ mg (sin θ – μ cos θ) x = \(\frac{1}{2}\) kx2
∴ 2 mg (sin 37° – μ cos 37°) = kx
⇒ 2 × 1 × 9.8 ms-2 (sin37°- μ cos37°) = 100 × 0.1m
⇒ 19.6 (0.601 – μ × 0.798) = 10
0.601 – μ × 0.798 = 0.5102
⇒ μ × 0.798 = 0.09079
∴ μ = \(\frac{0.09079}{0.798}\)= 0.1137