A box of a mass 2 kg is released from rest at the top of the ramp shown int the

Question

A box of a mass 2 kg is released from rest at the top of the ramp shown int the figure. The numbered circles indicate relevant situations for the problem:

(1) when the box is at the top of the ramp,

(2) when the box is at the bottom of the ramp(we assume a smooth transition from the ramp to the horizontal floor),

(3) at a distance s from the end of the ramp.

The friction coefficient of the ramp is 0.1, the friction coefficient of the level surface is 0.05, and we use g=10 m/s^2.

A) Calculate the work done by the friction force as the box slides down the ramp. The work due to friction is

1) -23J

2) -20J

3) -12J

4) 20 J

5) 120J

Explanation / Answer

The first thing we need to find is d.

d^2 = 6^2 + 10^2
d= 11.66m

Friction acts the entire way down the ramp. and the force of friction is given by:
Ff = FN

So we need to find the normal force. In this case, the normal force is found by calculating the component of gravity that is perpendicular to the plane:

Fgy = Fgcos where is the angle of the incline.

cos = 10m/11.66m = 0.858

So Fgy = mgcos = (2kg)(10m/s^2)(0.858) = 17.15N

This is also the magnitude of the normal force.

Force of friction is then

Ff = FN = 0.1(17.15N) = 1.715N

Therefore the magnitude of the work done by friction is:

Wf = Ff*d = (1.715N)(11.66m) = 20J

However, since friction acts in the opposite direction as the displacement and since friction always reduces the amount of kinetic energy, this value should be negative.

Wf = -20J

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