A Classical Approach of Newtonian Mechanics

5 CONSERVATION OF ENERGY 5.8 Power

m 50

The work WJ done by the frictional force is

WJ = −f x = −69.89 × 6 = −419.3 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 J 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 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 J
dissipated as heat. Hence,

K = W + WJ = 550.6 − 419.3 = 131.3 J.

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

v =

‚

., 2 K

=

‚

., 2 × 131.3

= 2.29 m/s.

Worked example 5.3: Stretching a spring

Question: The force required to slowly stretch a spring varies from 0 N to 105 N
as the spring is extended by 13 cm from its unstressed length. What is the force

constant of the spring? What work is done in stretching the spring? Assume that

the spring obeys Hooke’s law.

Answer: The force f that the spring exerts on whatever is stretching it is f = −k x,

where k is the force constant, and x is the extension of the spring. The minus
sign indicates that the force acts in the opposite direction to the extension. Since

the spring is stretched slowly, the force fJ which must be exerted on it is (almost)

equal and opposite to f. Hence, fJ = −f = k x. We are told that fJ = 105 N when

x = 0.13 m. It follows that

105

k =

0.13

= 807.7 N/m.