Problem 7-23 Work - Part 6 - 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

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

Correct!

\(W = mgh\) which is exactly the result we got before. This is because the only work done here is that in lifting the toboggan. Sideways motion requires no work since the friction is zero. If friction had been present, then only the first method would work taking into account the friction in evaluation \(F\).

\(W = mgh = (4.81\; kg)(9.8\; m/s^2)(27.3\; m) = 1.29\times 10^3 \; J\)

You have completed this problem.