Worked example 5.2: Dragging a treasure chest

Question: A pirate drags a 50kg treasure chest over the rough surface of a dock by exerting a constant force of 95N acting at an angle of $15^\circ$ above the horizontal. The chest moves 6m in a straight line, and the coefficient of kinetic friction between the chest and the dock is $0.15$. How much work does the pirate perform? How much energy is dissipated as heat via friction? What is the final velocity of the chest?

Answer: Referring to the diagram, the force $F$ exerted by the pirate can be resolved into a horizontal component $F \cos\theta$ and a vertical component $F \sin\theta$. Since the chest only moves horizontally, the vertical component of $F$ performs zero work. The work $W$ performed by the horizontal component is simply the magnitude of this component times the horizontal distance $x$ moved by the chest:

$$W = F \cos\theta x = 95\times \cos 15^\circ\times 6 = 550.6 {\rm J}.$$

The chest is subject to the following forces in the vertical direction: the downward force $m g$ due to gravity, the upward reaction force $R$ due to the dock, and the upward component $F \sin\theta$ of the force exerted by the pirate. Since the chest does not accelerate in the vertical direction, these forces must balance. Hence,

$$R = m g - F \sin\theta = 50\times 9.81 - 95\times\sin 15^\circ = 465.9 {\rm N}.$$

The frictional force $f$ is the product of the coefficient of kinetic friction $\mu_k$ and the normal reaction $R$, so

$$f = \mu_k R = 0.15\times 465.9 = 69.89 {\rm N}.$$

The work $W'$ done by the frictional force is

$$W' = - f x = - 69.89\times 6 = -419.3 {\rm J}.$$

Note that there is a minus sign in front of the $f$ because the displacement of the chest is in the opposite direction to the frictional force. The fact that $W'$ is negative indicates a loss of energy by the chest: this energy is dissipated as heat via friction. Hence, the dissipated energy is $419.3 {\rm J}$.

The final kinetic energy $K$ of the chest (assuming that it is initially at rest) is the difference between the work $W$ done by the pirate and the energy $-W'$ dissipated as heat. Hence,

$$K = W + W' = 550.6 - 419.3 = 131.3 {\rm J}.$$

Since $K= (1/2) m v^2$, the final velocity of the chest is

$$v = \sqrt{\frac{2 K}{m}} = \sqrt{\frac{2\times 131.3}{50}} = 2.29 {\rm m/s}.$$