Question

A 75-kg man pushes the same cart up a 10-m ramp elevated an angle of 15O. Find the work done.

Answer 1

I assume, he pushes the cart with constant velocity up along the ramp.

Then the force $\mathbf{F}$ applied to the cart by the men is equal to the friction force $\mathbf{F}_{fr}$ along the ramp and the projection of the gravitational force $m\mathbf{g}$ (see figure). Thus, obtain:

$$F = F_{fr} + mg\sin\alpha$$

where $m$ is the mass of the cart together with the man (assuming the task is to find all work done; if the task is to find the work done with respect to the cart, then $m$ will be the mass of the cart only), and $\alpha = 15\degree$ is the angle of inclination. By definition, the frictional force is:

$$F_{fr} = \mu N$$

where $\mu$ is the coefficient of friction, and $N = mg\cos \alpha$ is the normal force (see figure). Thus, obtain:

$$F = \mu mg\cos \alpha + mg\sin \alpha = mg(\mu \cos \alpha +\sin \alpha)$$

By difinition, the work is:

$$A = Fs$$

where $s = 10m$ is the length of the ramp. Finally, obtain:

$$A = smg(\mu \cos \alpha +\sin \alpha)$$

As one can see, there are several quantities, that are omitted in this prombem (namely, $m,\mu$). Thus, it is not possible to obtain a numerical answer. But the final formula was derived.

Answer. $A = smg(\mu \cos \alpha +\sin \alpha)$.