A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m⁻¹ as shown in the figure. The block is released from rest with spring in the unstructured position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and pulley is frictionless.
Answers
given, mass of block, m = 1kg
inclination angle , θ = 37°
spring constant, k = 100N/m
extension of spring, x = 10cm = 0.1m
see free body diagram,
normal reaction acting on body, R = mgcosθ
= 1 × 10 × cos37° = 10 × 4/5 = 8 N
so, frictional force acting on block, f = μR = μ (8) N
weight of body along plane , W = mgsinθ
= 1 × 10 × sin37° = 10 × 3/5 = 6 N
now from law of conservation of energy,
spring potential energy = workdone due to weight along plane + workdone due to frictional force
or, 1/2 kx² = W × x cos0° + f × x cos180° [frictional force just opposite direction of its motion. so, angle is 180° ]
or, 1/2 kx = W - f
or, 1/2 × 100 × 0.1 = 6 - μ(8)
or, 5 - 6 = -μ(8)
or, μ = 1/8 = 0.125
hence, coefficient of friction is 0.125
[note : Don't disappoint if answer is mismatched from your books or reference. just use value of g = 9.8 m/s². here I used 10m/s² because you didn't mention about g. generally people choose g = 10m/s² . that's why I also preferred. I hope you got it. cheers. ]
