Physics help?

Question

An elevator car, with a mass of 500 kg, starts at rest and moves upward with a constant acceleration until it reaches a velocity of 1.60 m/s after 2.70 seconds. If the car rises 2.16 m during this time, what is the average power of the elevator motor during this period? Ignore friction and other resistive forces, and the mass of the cables.

Answer 1

There are two obvious methods one could use to solve this problem, both giving the same result.

Remember that Power is the rate at which energy is used with respect to time.
Power = Energy / time

In this case, the energy we are talking about is the energy of the elevator after it has finished accelerating, which is also equal to the work done on the elevator.

Method 1)
The elevator experiences a change in velocity. With this change in velocity, the elevator experiences a change in Kinetic Energy. Additionally, the elevator’s gravitational Potential Energy increases. The change Kinetic Energy + the increase in PE equals the work done on the elevator. By dividing the change in the Energy by the time it took for the KE to change we can calculate the power.

KE = 1/2 mv^2
Where v is the speed of the elevator.

PE = mgh
Where m is the mass, g is the gravitational acceleration, and h is the distance the elevator is lifted up.

Since we know the elevator started from rest, the change in KE is equal to the elevator’s final KE.
m = 500 kg
v = 1.6 m/s
KE = 1/2 (500 kg) (1.6 m/s)^2
KE = 640 Joules

PE = mgh
PE = (500 kg) * (9.81 m/s^2) * (2.16 m)
PE = 10595

Change in Energy = change in KE + change in PE
Change in Energy = 11235

Power = Energy / time
Energy = 11235 J
Time = 2.70 second

Power = 11235 J / 2.70 s
Power = 4161 watts


Method 2)
The elevator is experiencing a constant acceleration, which means that a constant force is being applied to the elevator. From Newton’s 2nd law,
F = ma
We can find what this force is.

M = 500 kg
A = change in velocity / change in time = 1.6 m/s / 2.7 s = .5926 m/s^2
F = (500 kg) * (.5926 m/s^2)
F = 296.3 Newtons

To this value we must add in the force of gravity acting on the elevator as well which is equal to its weight.
Weight = mass * gravity
Weight = 500 kg * 9.81 m/s^2
Weight = 4905 Newtons

For a total force applied of,
F = 4905 N + 296.3 N = 5201.3 N

The work done on the elevator by a constant force (F) over some distance (d) is given as,
Work = F * d
We know F since we just calculated it and we are given a value of d = 2.16 meters.
Work = (5201.3 N) * (2.16 m)
Work = 11235 Joules

This work is done over a period of 2.7 seconds as before, so again we get a power of,
Power = 11235 J / 2.7 s
Power = 4161 watts

Answer 2

Power is equal to work divided by time. In this problem, you've been given the time, 2.70 seconds, so you need to find the work. Work can be defined as the amount of energy that was added to the system. The elevator had an increase in gravitational potential energy and an increase in kinetic energy. The elevator's speed increased from zero to 1.6 m/s, so its kinetic energy increased by 0.5*mv^2 = 0.5(500 kg)(1.6 m/s)^2 = 640 J. The gravitational potential energy increased by mgh = (500 kg)(9.8 m/s^2)(2.16 m) = 10584 J. The total energy increase, and thus work done by the motor, is 640 J + 10584 J = 11224 J, meaning the power is 11224 J / 2.70 s = 4157 W.

Answer 3

v(0) =0, v=1.60 m/s, t = 2.70 s, S=2.16 m and M=500 kg
Using v = v(0) + at , we get 1.60 =a* 2.70, which leads to constant acceleration:
a= 0.5926 m/s^2.
Force F on the elevator :
F=Mg+Ma =500 *9.8 + 500*0.5926=4900+296.3 N = 5196.3 N
Work W done by the motor on the elevator :
W= F*S = 5196.3 *2.16 = 11224 Joules
The power P of the motor:
P=W/t = 11224/2.70 =4157 W Answer.

Answer 4

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