class 051021
Do the problems you can. On the ones you can't figure out, refer to the Nutshell Summary and tell me what things on the list seem to apply to each problem, and as much as you can about how those things apply.
1. What is your approximate mass in kg? If you don't know, use 150 kg, but nobody in here has that much mass.
How much energy would it take for you to climb a ladder 6 meters high?
work/energy is calculated using F * `ds.
We need F and `ds.
`ds is 6 meters.
F is the force you have to exert to overcome gravity as your vertical position changes by 6 m.
This force is the weight m g = 150 kg * 9.8 m/s^2 = 1470 N.
You have to exert an upward force of 1470 N to climb the ladder.
Your force and displacement are both upward, so using upward as the positive direction:
F * `ds = 1470 N * 6 m = 9,000 N m = 9,000 J.
By how much would your gravitational PE change in the process?
The 9000 J was done against the conservative force of gravity, and the definition of PE is work done against a conservative force. So `dPE_grav is 9000 J.
If you wanted to use the formula `dPE_grav = m g `dy you would get
`dPE_grav = 150 kg * 9.8 m/s^2 * 6 m = 9000 J approx.
This is identical to the calculation done previously.
2. How much work would be required to stop a 1200 kg automobile originally moving at 20 m/s?
The most basic form of the work-energy theorem tells us that `dW_net_ON = `dKE. We have the information to find `dKE (see below) and we easily determine that `dKE = -240,000 J.
So `dW_net_ON = -240,000 J.
This is the work done by the net force acting on the automobile. The net force acting on the automobile to stop it is in one direction and the displacement of the automobile is in the other, so the net force and the displacement are in opposite directions, making the work negative.
By how much does the KE of this automobile change as it stops?
We can use the definition of KE as 1/2 m v^2 to find initial and final KE, and therefore change in KE.
KE0 = 1/2 * 1200 kg * (20 m/s)^2 = 240,000 kg m^2 / s^2 = 240,000 J.
KEf = 0, since vf = 0.
So `dKE = -240,000 J.
3. Two 100-kg football players traveling in opposite directions, each at 5 m/s, collide and come to rest.
What is their total KE change?
Each 100-kg mass is initially moving at 5 m/s and hence has KE equal to 1/2 ( 100 kg ) * (5 m/s)^2 = 1250 Joules.
The total KE of the two masses is therefore 2500 Joules.
After collision the masses are at rest and therefore have KE equal to 0.
So the total KE change is KEf - KE0 = 0 - 2500 J = -2500 J.
What happens to the KE lost?
The lost KE is ultimately converted to very low-grade thermal energy (thermal energy we can't recover in any useful form, even in principle).
4. If I exert a uniform 40-Newton force on a ball of mass .6 kg through a throwing motion that covers a distance of 1.5 meters, then what speed will the ball attain? Solve this problem using the work-energy theorem.
Assuming no other significant forces acting in the direction of motion, the net force will be 40 N.
The work done by that net force is therefore
`dW_net_ON = 40 N * 1.5 m = 60 Joules.
This work is done on the ball.
Since `dW_net_ON = `dKE we see that `dKE must be 60 Joules.
It's also reasonable to assume that the ball started from rest, so the change in KE is just the final KE, and final KE is 60 Joules.
Final KE is 1/2 m vf^2 so
1/2 m vf^2 = 60 J
vf = +- sqrt( 2 * 60 J / m) = +- sqrt(2 * 60 J / (.6 kg) ) = +- sqrt(200 m^2 / s^2) = +- 14 m/s, approx.
On the force vectors experiment, for each setup:
Note that your forces will be in units of dominoes. We will be able to convert dominoes to Newtons, but right now you don't know what the conversion is, so just keep forces in units of dominoes.