Question

A 200-g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60° above the horizontal. Determine how far up the incline the block moves from its initial position before it stops if the coefficient of kinetic friction is 0.400.

Answer 1

Answer:

L = 3.35 m

Explanation:

given,

mass of the block = 0.2 Kg

spring constant = 1.40 kN/m

until spring compressed = 10 cm

inclination of ramp = 60°

now,

initial potential energy of spring = (work done by friction) + (final gravitational potential energy of block)

$\dfrac{kx^2}{2} = \mu NL + mg y$

$\dfrac{kx^2}{2} = \mu mg cos \theta + mg L sin \theta$

$L = \dfrac{kx^2}{2mg(\mu\ cos \theta + sin \theta)}$

$L = \dfrac{1400\times 0.1^2}{2\times 0.2 \times 9.8(0.4\ cos 60^0 + sin 60^0)}$

L = 3.35 m