Question

A small box with mass 0.600 kg is placed against a compressed spring at the bottom of an incline that slopes upward at 37.0∘ above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is μk=0.400. The spring is released and the box travels up the incline, leaving the spring behind.

What minimum elastic potential energy must be stored initially in the spring if the box is to travel 2.00 m from its initial position to the top of the incline?

Answer 1

Answer:

$10.845\ \text{J}$

Explanation:

m = Mass of box = 0.6 kg

$\theta$ = Angle of incline = $37^{\circ}$

$\mu_k$ = Coefficient of kinetic friction = 0.4

l = Distance the box travels = 2 m

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

The potential energy in the spring is given by

$U=mgl\sin\theta+mg\mu_k\cos\theta l\\\Rightarrow U=mgl(\sin\theta+\mu_k\cos\theta)\\\Rightarrow U=0.6\times 9.81\times 2\times(\sin37^{\circ}+0.4\times\cos37^{\circ})\\\Rightarrow U=10.845\ \text{J}$

The minimum elastic potential energy required is $10.845\ \text{J}$.