Question

OR
A 1 kg block situated on a rough incline is connected to a spring of spring const
pelow. The block is released from rest with the spring in the unstretched position.
down the incline before coming to rest. Find the coefficient of friction between the
assume that the spring has a negligible mass and the pulley is frictionless.
lected to a spring of spring constant 100 Nm-1 as shown
le spring in the unstretched position. The block moves 10 cm
e coefficient of friction between the block and the incline
K= 100 Nm"
ത്ത
i Kg
30°​

Answer 1

Explanation:

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°)

At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,

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

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