Problem 7-23 Work - Part 5 - C

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

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

Correct!

The force in the vertical direction is equal to \(mg\) just the force required to lift the toboggan.

The work done by the vertical force is:

(A)   \(mgh\)

(B)   \(mgd\)

(C)   \(mg(d - h)\)