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)
[tex]\dfrac{kx^2}{2} = \mu NL + mg y[/tex]
[tex]\dfrac{kx^2}{2} = \mu mg cos \theta + mg L sin \theta [/tex]
[tex]L = \dfrac{kx^2}{2mg(\mu\ cos \theta + sin \theta)}[/tex]
[tex]L = \dfrac{1400\times 0.1^2}{2\times 0.2 \times 9.8(0.4\ cos 60^0 + sin 60^0)}[/tex]
L = 3.35 m
