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## 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 above the horizontal. The chest moves 6m in a straight line, and the coefficient of kinetic friction between the chest and the dock is . 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 exerted by the pirate can be resolved into a horizontal component and a vertical component . Since the chest only moves horizontally, the vertical component of performs zero work. The work performed by the horizontal component is simply the magnitude of this component times the horizontal distance moved by the chest:

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

The frictional force is the product of the coefficient of kinetic friction and the normal reaction , so

The work done by the frictional force is

Note that there is a minus sign in front of the because the displacement of the chest is in the opposite direction to the frictional force. The fact that is negative indicates a loss of energy by the chest: this energy is dissipated as heat via friction. Hence, the dissipated energy is .

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

Since , the final velocity of the chest is

Next: Worked example 5.3: Stretching Up: Conservation of energy Previous: Worked example 5.1: Bucket
Richard Fitzpatrick 2006-02-02