## Examples

A spring with spring constant k = 500 N/m is compressed a distance x = 0.10 m. A block of mass m = 0.250 kg is placed next to the spring, sitting on a frictionless, horizontal plane. When the spring and block are released, how fast is the block moving as it leaves the spring?When the spring is compressed, work is required and the spring gains potential energy,

U

_{s}= (^{1}/_{2}) k x^{2}U

_{s}= (^{1}/_{2}) (500 N/m) (0.10 m)^{2}U

_{s}= 2.5 JAs the spring and mass are released, this amount of

workdoneto the massto change its kinetic energy from zero to a final value ofK = (

^{1}/_{2}) m v^{2}K = (

^{1}/_{2}) (0.25 kg) v^{2}K = (

^{1}/_{2}) (0.25 kg) v^{2}= 2.5 J = U_{s}= W_{s}v

^{2}= 20 m^{2}/ s^{2}v = 4.47 m/s

Explain the actions of a pole vaulter in terms of Energy Conservation.As the pole vaulter

runstoward the bar, he gainskinetic energy.As hebendsthe pole, he doesworkon the pole and storeselastic potential energyin the pole -- this is much like compressing a spring or pulling a bow. That stored elastic potential energy is converted intogravitational potential energyas he goes up and over the bar.An interesting sidelight: A good pole vaulter -- or high jumper -- will "curl" himself or herself "around" the bar so that his or her

center of massmay gounderthe bar so less energy is required.

**A skier starts from rest at the top of a
frictionless incline of heigh 20 m as shown here. At the bottom of
the incline, the skier encounters a horizontal surface where the
coefficient of kinetic friction between the skis and snow is 0.210.
How far does the skier travel on the horizontal surface before coming
to rest?**

While coming

downthe incline, energy isconserved.We will measure gravitational potential energywith respect tothe horizontal surface, so U_{gf}= 0. The skier starts from rest, so K_{i}= 0 . ThenE

_{i}= E_{f}K

_{i}+ U_{gi}= K_{f}+ U_{gf}0 + m g h

_{i}= (^{1}/_{2}) m v_{f}^{2}+ 0m (9.8 m/s

^{2}) (20 m) = (1/2) m v_{f}^{2}v

_{f}^{2}= 392 m^{2}/ s^{2}v

_{f}= 19.8 m / sAs the skier moves across the "rough snow" on the horizontal plane, friction does (negative) work and reduces the skier's Kinetic Energy to zero when the skier stops. This work done by friction is

W

_{f}= - F_{f}sThe friction force F

_{f}is given byF

_{f}= F_{n}On this horizontal surface, it turns out that F

_{n}= m gF

_{f}= F_{n}= = m g = (0.210) m (9.8 m/s^{2}) = (2.06 m/s^{2}) m(Be careful, one of these m's stands for meters, as in m/s

^{2}, while the other m is themassof the skier! And we are also using s for seconds, as in m/s^{2}, and also for the distance moved!)This means the work done by friction is

W

_{f}= - F_{f}s = - [ (2.06 m/s^{2}) m ] sAt the beginning of this horizontal section, the skier's original Kinetic Energy is

K

_{o}= (^{1}/_{2}) m v^{2}= (^{1}/_{2}) m (19.8 m / s)^{2}= 196 (m/s)^{2}mWhen the skier comes to rest, the final Kinetic Energy will be zero,

K

_{f}= 0The

onlywork done is the work due to friction so that is thenetwork. And we know thenet workequals thechangein the Kinetic Energy,W

_{f}= K = K_{f}- K_{o}= 0 - 196 (m/s)^{2}m- [ (2.06 m/s

^{2}) m ] s = - 196 (m/s)^{2}m[ (2.06 m/s

^{2})] s = 196 (m/s)^{2}s = 95.15 m

[[ Be careful! We have used m to mean the

massof the skiier and we have also used m to meanmeters. And s is used for two different meanings, too! We have used s to meandistanceand we have also used s to meanseconds. Just watch the details! ]]

(c) Doug Davis, 2001; all rights reserved