Power needed to push an object vertical at constant velocity

In summary: Correct-o. Can you figure out then what the force exerted by the motor is?The force exerted by the motor must be equal to the force of gravity, which is equal to the weight of the ride. So the force exerted by the motor is 1.20×105 * 9.8 = 1,176,000 N.Now we can calculate the power:P = F * vP = (1,176,000)(3.96)P = 4,654,560 wattsIn summary, the minimum power rating of the motor needed to raise the 1.20×105kg ride from the ground to the top in 10.0 seconds at a constant velocity of
  • #1
IAmPat
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0

Homework Statement



Stuntman's Freefall, a ride at Six Flags Great Adventure in NewJersey, stands 39.6 meters high. Ignoring the force of friction, what is the minimum power rating of the motor that raises the 1.20×105kg ride from the ground to the top in 10.0 seconds at a constant velocity?

Homework Equations



Constant Velocity = 39.6 / 10 = 3.96m/s
1.20×105 = 120000.0 kg

The Attempt at a Solution



I've tried it two ways now, but clearly neither of them worked.

First way:

[tex]\Delta[/tex]KE = [tex]\frac{1}{2}[/tex] * (1.20×105) * (3.96)2
[tex]\Delta[/tex]KE = 940,896
W = [tex]\Delta[/tex]KE
W = 940,896

P = [tex]\frac{W}{t}[/tex]
P = [tex]\frac{940896}{10}[/tex]
P = 94089.6 watts

This was wrong... so I tried this way

2nd way:

Velocity = 3.96m/s
Acceleration = .396m/s2
[tex]\sum[/tex]Fy = (1.20×105) * .396m/s2
[tex]\sum[/tex]Fy = 47,520

W = F * [tex]\Delta[/tex]y
W = 47250 * 39.6
W = 1,881, 792

P = [tex]\frac{W}{t}[/tex]
P = [tex]\frac{1881792}{10}[/tex]
P = 188179.2 wattsRan out of ideas. What am I missing here?
 
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  • #2
Constant velocity means ΔKE = zero.

Try power P = Fv where F is the force supplied by the motor. Can you find F?
 
  • #3
kuruman said:
Constant velocity means ΔKE = zero.

Try power P = Fv where F is the force supplied by the motor. Can you find F?

Then I end up with the same answer as my second try though. Unless I'm calculating Force wrong?

Velocity = 3.96m/s
Acceleration = .396m/s2
Fy = m*a
Fy = (1.20×105) * .396m/s2
Fy = 47,520

P = F * v
P = (47520)(3.96)
P = 188179.2 watts
 
  • #4
You need to convert velocity to acceleration.
 
  • #5
Chronos said:
You need to convert velocity to acceleration.
I am not sure what this means. If the velocity is constant, the acceleration is zero. Here, we are neglecting friction, so there are only two forces acting on the ride, the lifting mechanism and gravity. If the ride moves at constant velocity, what must the force of the lifting mechanism be equal to?
 
  • #6
If sumF = 0 then a = 0. The only forces are gravity and the motor.
 
  • #7
Correct-o. Can you figure out then what the force exerted by the motor is?
 

What is the definition of power?

Power is the rate at which work is done or energy is transferred. It is usually measured in watts (W) or horsepower (hp).

How is power related to pushing an object vertically at constant velocity?

In order to push an object vertically at a constant velocity, a certain amount of power is needed to overcome the force of gravity. This power is determined by the weight of the object and the height it needs to be pushed.

What factors affect the amount of power needed to push an object vertically at constant velocity?

The weight of the object, the height it needs to be pushed, and the force of gravity are the main factors that affect the amount of power needed. Other factors such as friction and air resistance may also play a role.

How can the power needed to push an object vertically at constant velocity be calculated?

The formula for power is P = W/t, where P is power, W is work, and t is time. In the case of pushing an object vertically at constant velocity, the work done is equal to the force of gravity multiplied by the distance the object is pushed, and the time is the duration of the push.

What are some real-life examples of pushing an object vertically at constant velocity and the power needed?

Some examples of pushing an object vertically at constant velocity and the power needed include using an elevator, lifting objects onto a shelf, and pushing a cart up a ramp. The amount of power needed will vary depending on the weight of the object and the distance it needs to be pushed.

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