#$&*
course phy 201
11/11 11:30 pm
`q001. In today's lab activity you found the work done by the rubber band used in the preceding class to energize the rotating ramp.
Give your data.
****
Mass held by length of rubber band chain: 9 dominoes; change in rubber band chain length: 22 cm; 99 cm/domino weights
#$&*
Show how you analyzed your data.
@&
I understand
9 dominoes; change in rubber band chain length: 22 cm
I don't know what you mean by 99 cm/domino weights or how you arrived at that result.
*@
@&
From your units it would appear that you divided some quantity in cm by some quantity in domino weights.
*@
****
By forming a graph of domino weights vs. rubber band length, we found a set triangle of data. We then found the area of this triangle.
#$&*
Give your conclusions.
****
The work done by a rubber band is proportional to where the rubber band’s force is applied.
@&
If the rubber band has the same change in length, then it does the same work. The average force and the displacement would be the same no matter where it was applied.
*@
#$&*
If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
****
#$&*
`q002. Using the moment of inertia of the ramp and dominoes (found in the preceding document) and the energy in your rubber band chain, find the angular velocity of the rotating ramp, assuming that all of the potential energy stored in the rubber band chain is transferred to the ramp and that none of that energy has yet been dissipated by friction.
****
moment of inertia: 80,000 grams cm^2; energy: 99 cm^2/seconds^2 domino weights;
@&
It's not clear how you got this. The units of this calculation are length^2 / seconds^2 * force. The units of work are the units of length * force. So your units here are not units of work.
*@
KE = 1/2I * angular velocity^2 99 cm^2 domino weights/seconds^2 = 80 kg cm^2 * angular velocity^2; assume one domino weight = 16 grams;
@&
16 grams is a mass, not a weight
*@
1.58 kg cm^2/s^2; 1.58 kg cm^2/s^2 = 40 kg cm^2 * angular velocity^2; 0.0395 /s^2 = angular velocity^2; 0.1987 radians/s = angular velocity; roughly 11.39 degrees/second;
@&
Your reasoning appears pretty good, but your quantities are not all correct.
*@
#$&*
`q003. If my truck has mass 1500 kg and descends through a total rise of magnitude 25 meters while slowing from 50 mph to 40 mph, by how much does its kinetic energy change, and how much work is done on it by the conservative gravitational force?
****
gravitational force: 14700 newtons; distance travelled: 25 meters; work done: 367500 joules; KE = ½ m * v^2; initial KE: 1,875,000; final KE: 1,200,000; change in KE: - 675,000 joules;
@&
You don't have units on your kinetic energies and don't show how they were calculated.
My guess is that the kinetic energies you report are in kg * miles^2 / hour^2, not in Joules.
*@
#$&*
How much work do nonconservative forces do on the truck?
****
to find actual acceleration; vf^2 = v0^2 + 2a * ds; 40^2 = 50^2 + 2a * 25; 1600 = 2500 + 2a * 25; - 900 = 2a * 25; - 36 = 2a; -18 m/s^2 = a; acceleration = -18 m/s^2; applied force: -27000 newtons; -27000; work = force * displacement: -27000 * 25 = -675,000 joules;
@&
Your calculations are mixing up miles / hour with distances in meters. You need to show units throughout any calculation. If you did so on this series of calculations you would see that your units don't mesh, and would thereby be reminded to do the appropriate conversions.
*@
#$&*
Road friction exerts a force equal to about 2% of the truck's weight. If it coasted a distance of 1000 meters along the road, what is the work done on the truck by friction?
****
2% of 14,700 newtons; 294 newtons of friction; work done by friction: 294,000 joules;
#$&*
Friction is one of the nonconservative forces acting on the truck. What is the work done on the truck by all other nonconservative forces combined?
****
-675,000 joules + 294,000 joules = -381,000 joules;
@&
Your -675 000 Joules isn't right, but your 294 000 Joules and your overall idea are.
*@
#$&*
What average force did those other nonconservative forces have to exert during the truck's 1000 meter displacement?
****
-15,240 newtons;
#$&*
My 2000 kg station wagon coasts along the same path and its speed decreases from 50 mph to 47 mph. Friction again exerts a force equal to 2% of the vehicle's weight. What average force was exerted by nonconservative forces other than friction?
****
2209 = 2500 + 2a * 25; -291 + 2a * 25; - 11.64 = 2a; -5.82 = a; force: -11,640 newtons; nonconservative force: -11,640 newtons;
#$&*
"
@&
Your thinking is mostly good. However you aren't using units and you are making serious errors as a result.
*@