Question

A 4.5 kg box slides down a 4.3-m-high frictionless hill, starting from rest, across a 2.0-m-wide horizontal surface, then hits a horizontal spring with spring constant 510 N/m. The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.0-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.22. What is the speed of the box just before reaching the rough surface? What is the speed of the box just before hitting the spring? How far is the spring compressed? Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

Answer 1

Answer:

speed  before reaching rough surface = 9.18 m/s

speed before hitting spring = 8.70 m/s

spring compression = 82 cm

number of complete trip = 9

Explanation:

Lets say

Position 1: On top of hill

Position 2: down the hill

Position 3: after the rough surface

Position 4: after hitting the spring

We'll strictly use conservation of energy for this equation

Potential energy on top of energy is full converted into kinetic energy down the hill (since surface is frictionless)

Hence, PEg1 = KE2

mgh = (1/2)mv2^2

(4.5)(9.8)(4.3) = (1/2)(4.5)v2^2

189.63 = (1/2)(4.5)v2^2

v2^2 = 2(9.8)(4.3) = 84.28

v2 = sqrt(84.28) = 9.18 m/s

After down the hill, it passes a rough surface. So some of the energy is loss due to friction forces

Friction force, Ff = u (coeff of kinetic friction ) x N (normal force)

Normal force, N = weight of box = mg = 4.5 x 9.8

Ff = 0.22 x 4.5 x 9.8

Work done / Energy loss = Wf = Ff x d (distance)

Wf = 0.22 x 4.5 x 9.8 x 2 = 19.404

Energy after passing the rough surface is totally kinetic energy

KE3 = KE2 - Wf = 189.63 - 19.404 = 170.226

speed after rough surface,

(1/2)mv3^2 = 170.226

v3 = sqrt((2 x 170.226)/4.5) = 8.70 m/s

After hitting the spring, all the kinetic energy is converted into potential energy of spring

170.226 = (1/2)kx^2

x^2 = 2 x 170.226 / 510     {note that constant of spring, k = 510}

x^2 = 0.668

x = sqrt(0.668) = 0.82m (82 cm)

To calculate complete trip before the box coming to rest, note that the only place where it loss energy is at the rough surface.

Energy before the first time pass rough surface = 189.63

Energy loss each time passing rough surface = 19.404

189.63 / 19.404 = 9.773 (9 complete with balance of 0.773)

That mean, the box will pass the rough surface 9 complete trip before coming to rest