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, PEel = (1/2) k x2
PEel = (1/2) (500 N/m) (0.10 m)2
PEel = 2.5 J
As the spring and mass are released, this amount of work is done to the mass to change its kinetic energy from zero to a final value of
K = (1/2) m v2
K = (1/2) (0.25 kg) v2
K = (1/2) (0.25 kg) v2 = 2.5 J = PEel = Ws
v2 = 20 m2 / s2
v = 4.47 m/s
Explain the actions of a pole vaulter in terms of Energy Conservation.
As the pole vaulter runs toward the bar, he gains kinetic energy. As he bends the pole, he does work on the pole and stores elastic potential energy in the pole -- this is much like compressing a spring or pulling a bow. That stored elastic potential energy is converted into gravitational potential energy as 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 mass may go under the 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 down the incline, energy is conserved. We will measure gravitational potential energy with respect to the horizontal surface, so PEg,f = 0. The skier starts from rest, so Ki = 0 . Then
Ei = Ef
Ki + PEel ,i = Kf + PEel ,f
0 + m g hi = (1/2) m vf2 + 0
m (9.8 m/s2) (20 m) = (1/2) m vf2
vf2 = 392 m2 / s2
vf = 19.8 m / s
As 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
Wf = - Ff s
The friction force Ff is given by
Ff = Fn
where Fn is the normal force (perpendicular to the surface) On this horizontal surface, it turns out that Fn = m g
Ff = Fn = = m g = (0.210) m (9.8 m/s2) = (2.06 m/s2) m
(Be careful, one of these m's stands for meters, as in m/s2, while the other m is the mass of the skier! And we are also using s for seconds, as in m/s2, and also for the distance moved!)
This means the work done by friction is
Wf = - Ff s = - [ (2.06 m/s2) m ] s
At the beginning of this horizontal section, the skier's original Kinetic Energy is
Ko = (1/2) m v2 = (1/2) m (19.8 m / s)2 = 196 (m/s)2 m
When the skier comes to rest, the final Kinetic Energy will be zero,
Kf = 0
The only work done is the work due to friction so that is the net work. And we know the net work equals the change in the Kinetic Energy,
Wf = K = Kf - Ko = 0 - 196 (m/s)2 m
- [ (2.06 m/s2) m ] s = - 196 (m/s)2 m
[ (2.06 m/s2)] s = 196 (m/s)2
s = 95.15 m/s
(c) 2002, Doug Davis; all rights reserved
Nonconservative Forces Summary Return to ToC, Work and Energy