class 051014
Homework:
All students work Intro Problem Set problems in Set 5, Problems 1-6.
Read Chapter 6
Homework Problems:
Phy 121: Problems 1, 2, 3, 4, 16, 17, 26, 27, 34, 35, 47, 53, 58, 61
Phy 201: Problems 19, 22, 30, 32, 33, 38, 39, 41, 42, 45, 49, 50, 53, 61, 65
051014
What is the PE change of a 5 kg object as it slides down an incline of length 20 meters, on which the vertical drop is 10 meters? Assume no conservative forces act other than gravity.
Quick answer:
The change in PE is equal to the weight times the change in vertical position, which is
`dPE = (5 kg * 9.8 m/s^2) * (-10 m) = -490 N.
Complete reasoned out solution in terms of the definitions:
PE change is the work done by the system against the gravitational force.
When working with gravitational PE near the surface of the Earth, it's best to consistently use 'up' as the positive direction.
The work done by gravity on the system is weight * vertical displacement = -(5 kg * 9.8 m/s^2) * (-10 m) = 490 kg m^2 / s^2 = 490 Joules.
So the PE change being the negative of the work done by gravity on the system will be -490 Joules.
If friction exerts a force of 3 N on this incline, then how much work is done by this force?
Assuming down the incline to be the positive direction, the frictional force is -3 N and the displacement is 20 m so the work done by friction on the system is
`dW_frict_On = -3 N * 20 m = -60 J.
If the object has a velocity of 4 m/s at the top of the incline, then what is its KE at this point?
KE = 1/2 m v^2 and v = 4 m/s so we have KE = 1/2 * 5 kg * (4 m/s)^2 = 40 kg m^2 / s^2 = 40 Joules.
If the object gains 100 Joules of kinetic energy on the incline, then what is its final KE?
We just saw that the KE at the top is 40 Joules, and the gain is 100 Joules so the KE at the bottom will be 40 J + 100 J = 140 J.
Symbolically we have
KEf = KE0 + `dKE = 40 J + 100 J = 140 J.
What therefore is its final velocity?
KEf = 1/2 m vf^2 so
vf = +-sqrt(2 KE / m) =
+- sqrt(2 * 140 J / (5 kg) ) =
+- sqrt( (56 kg m^s / s^2) / (5 kg) ) =
+- sqrt( 56 m^2 / s^2) =
+-7.5 m/s.
Since downward is the positive direction and we know that the object will be moving downward, we conclude that
vf = +7.5 m/s.
In the process described above:
We found that -60 J is done ON the system by friction.
Since -60 J was done ON the system by friction, -(-60 J) = +60 J was done BY the system against friction.
We found that gravity did +490 J of work ON the system.
The work done BY the system against gravity is -490 J.
How does all this fit into the law of conservation of mechanical energy?
`dKE + `dPE + `dW_nc_BY = 0 so
+100 J - 490 J + `dW_nc_BY = 0 so that
`dW_nc_BY = +390 J.
`dW_frict_BY = +60 J. So there are an additional 330 J of work being done BY the system against other nonconservative forces. The nature of the force doing this work isn't specified, except that it's nonconservative because the PE change takes care of all the conservative forces. Maybe wind resistance, maybe someone trying to hold it back.