Energy conservation and transfer
Problem#1:
A 100 kg car in a roller coaster has total of 30,000 Joules of gravitational energy at point A. The car moves down a frictionless track and comes to a stop as it compresses a spring at point D.
1) Quantitatively sketch the energy bar graphs when the car is at the indicated points (the letters above: A, B, C, D).
2) How many Joules of kinetic energy does the car have at point C?
3) How fast is the car moving at point B?
2) How many Joules of kinetic energy does the car have at point C?
3) How fast is the car moving at point B?
Energy Bar graphs*:
*double click to enlarge
The graph at point A indicates that the only energy stored is in Ug, which is the gravitational energy stored due to the height of the car, point B has a relatively smaller Ug, because it is farther down the track than point A; also, K is storing some power in the velocity of the car moving down the track. At point C,a ll of the energy is stored into K (the velocity of the car). Point D has energy stored in both Ug and Us because the car has compressed the spring to create tension, and it is slightly above the ground.
Equations and explanations:
2)
K = 1/2m*v
K = 1/2(100kg * 600 m/s)
K = 30,000 J
At point C there has to be 30,000 J because there is not any energy stored in either Ug (height) or Us (spring constant).
3)
Ug = m*g*(height in meters)
Ug = (100 kg*9.8m/s^2*10m)
Ug = 9800 J
K = 30,000 J — 9800 J
K = 20,200 J
To calculate this we must first calculate the Ug at point B because we don't know the velocity at that point to be able to find K. Then we must subtract 9800 J form the total possible amount, which would yield us K.
K = 1/2 m*v^2
20,200 J = 1/2 *100kg*v^2
V^(1/2) = 202^(1/2)
V = 14.2 m/s
Next we have to solve the equation for K for the variable V, which is the velocity. By plugging in each number that is known, we can solve for V.
Therefore, the velocity is 14.2 m/s.
K = 1/2m*v
K = 1/2(100kg * 600 m/s)
K = 30,000 J
At point C there has to be 30,000 J because there is not any energy stored in either Ug (height) or Us (spring constant).
3)
Ug = m*g*(height in meters)
Ug = (100 kg*9.8m/s^2*10m)
Ug = 9800 J
K = 30,000 J — 9800 J
K = 20,200 J
To calculate this we must first calculate the Ug at point B because we don't know the velocity at that point to be able to find K. Then we must subtract 9800 J form the total possible amount, which would yield us K.
K = 1/2 m*v^2
20,200 J = 1/2 *100kg*v^2
V^(1/2) = 202^(1/2)
V = 14.2 m/s
Next we have to solve the equation for K for the variable V, which is the velocity. By plugging in each number that is known, we can solve for V.
Therefore, the velocity is 14.2 m/s.
Problem#2:
Adam and Steve challenge each other to a dash up a set of stairs (Δh = 4.0 m). Adam weighs 120 kg and makes the climb in 3.0 seconds. Steve weighs 150 kg and it took him 4.0 seconds to make it to the top.
1) Who has more Eg at the top?
2) Who, of the two, is more powerful?
1) Who has more Eg at the top?
2) Who, of the two, is more powerful?
equations and explanations:
1) First, I must solve for both men's Ug, which will allow me to determine their Eg.
Adam:
Ug = m*g*Δx
Ug = (120 kg)(9.8 m/s)(4.0 m)
Ug = 4704 J
Steve:
Ug = m*g*Δx
Ug = (150 kg)(9.8 m/s)(4.0 m)
Ug = 5880 J
Therefore, Steve has more Eg because he has a larger mass than Adam; Eg is the total amount of energy stored in a given situation.
Adam:
Ug = m*g*Δx
Ug = (120 kg)(9.8 m/s)(4.0 m)
Ug = 4704 J
Steve:
Ug = m*g*Δx
Ug = (150 kg)(9.8 m/s)(4.0 m)
Ug = 5880 J
Therefore, Steve has more Eg because he has a larger mass than Adam; Eg is the total amount of energy stored in a given situation.
2) To solve for the second half of the problem, I must first calculate the Force that each man exerted. The force will then allow me to calculate the weight of both men by plugging in the force. I would then continue to solve for P in the equation P = w/t which would then give me the power of each man, therefore the answer. With all of the numbers in these equations, it's just a simple plug in with equations:
F = m*v
W = N*t*cos45°
P = W/t
Steve:
F = 150 kg*1m/s = 150 N
W = 150(4)cos45° = 424 W
P = 424/4 = 106.1 J
Adam:
F = 120 kg* 4m/3s = 160 N
W = 160(4)cos45° = 452 W
P = 452/3 = 150.8 J
Therefore, Adam was the most powerful of the two.
F = m*v
W = N*t*cos45°
P = W/t
Steve:
F = 150 kg*1m/s = 150 N
W = 150(4)cos45° = 424 W
P = 424/4 = 106.1 J
Adam:
F = 120 kg* 4m/3s = 160 N
W = 160(4)cos45° = 452 W
P = 452/3 = 150.8 J
Therefore, Adam was the most powerful of the two.