Question

A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is 0.19.
Calculate the work done by the applied horizontal force.
Calculate the work done by the weight of the trunk.
How much energy was dissipated by the frictional force acting on the trunk?
From now suppose the 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a force along the plane (not as in the figure). The coefficient of kinetic friction between the trunk and the incline is 0.19.
Calculate the work done by the applied force.
Calculate the work done by the weight of the trunk.
How much energy was dissipated by the frictional force acting on the trunk?

Answer 1

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

$F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)$

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

$0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N$

$F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N$

$W_{F_x} = F_xd = 227.5\times 4.8 = 1092J$

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

$W_g = -mg\sin(30^\circ)d$

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

$W_g = -823J$

The energy dissipated by the frictional force can be found as follows:

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

$0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N$

$W_F = Fd = 227.5\times 4.8 = 1092J$

$W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J$

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.