Problem 7-23 Work - Part 4 - A

A girl pulls a toboggan of mass \(4.81 \;kg\) up a hill inclined at \(25.7^\circ\) to the horizontal. The vertical height of the hill is \(27.3 \;m\). Neglecting friction between the toboggan and the snow, determine how much work the girl must do on the toboggan to pull it at constant velocity up the hill. [Ans. \(1.29\times 10^3\; J\)]

Accumulated Solution

\(F = mg \sin \theta \\ d = h/ \sin \theta\)

Correct!

\(W = Fd = mg \sin \theta \times \bigg( \frac{h}{sin \theta} \bigg) = mgh\)

Notice that all reference to the angle cancels out. In other words this is the answer for a hill of any angle (so long as there is no friction).

This indicates the second attack on the problem.

The force acting on the toboggan in the vertical direction is given by:

(A)   \(N = mg\)

(B)  \(N \sin \theta + F \cos \theta = mg\)

(C)  \(N \cos \theta + F \sin \theta = mg\)