Question

Question 6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. 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. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Chapter Work, Energy And Power Page 138

Answer 1

Given,
Force constant (K) = 100 N/m
angle of inclinations = 37°
break in components of weight (mg )
in equilibrium ,
Normal reaction of the inclined plane ( N) = mgcos∅

we know,
friction = uN
where u is coefficient of friction .
friction = umgcos∅ .
hence, net Force acting along inclined plane = mgsin∅ - fr
= mgsin∅ - umgcos∅

in equilibrium condition ,
work done = PE of stretched spring
net force x displacement = 1/2kx²
[ use formula , PE = 1/2 Kx² ]
(mgsin∅ - umgcos∅)× x = 1/2Kx²
here,
m = 1 Kg
K = 100 N/m
x = 10cm
g = 10 m/s²
∅ = 37°

u = (2mgsin37° - kx)/2mgcos37°
= ( 2×1×10×3/5 - -100×0.1)/2×1×10×4/5
= (12 - 10)/16
= 2/16
= 1/8
u = 1/8
hence ,coefficient of friction = 1/8

Answer 2

Mass of the block, m = 1 kg
Spring constant, k = 100 N m–1
Displacement in the block, x = 10 cm = 0.1 m


At equilibrium:
Normal reaction, R = mg cos 37°

Frictional force, f = μ R = mg Sin 370

Where, μ is the coefficient of friction
Net force acting on the block = mg sin 37° – f
= mgsin 37° – μmgcos 37°

= mg(sin 37° – μcos 37°)


mg(sin 37° – μcos 37°)x = (1/2)kx2

1 × 9.8 (Sin 370 - μcos 37°) = (1/2) × 100 × (0.1)

0.602 - μ × 0.799 = 0.510

∴ μ = 0.092 / 0.799 = 0.115