# Calculate power with v on inclined plane with miu

• sciencescienc
In summary, The conversation involves a person seeking help in solving a physics problem involving dragging an object down an inclined plane. The conversation includes equations and attempts at finding the correct answer until the correct solution is eventually found. The summary also notes the importance of considering net forces and the role of friction in the problem.
sciencescienc

## Homework Statement

please advise how to solve. seems i can't without d given, or t!

a man drags a 130 kg object down an inclined plane at a speed of 1.40m/s
The coefficient of friction between the object and the inclined plane is 0.800 and the angle of inclination with the horizontal is 23 degrees. What is the amount of power this man expends in dragging the object?

## Homework Equations

Work = F d
Work = F cos theta d

Power = Work/time
Power = F v

Work = change in kinetic energy

## The Attempt at a Solution

F friction = 0.8 ( mg cos 23) = 95.7N
Power = F.v = 95.7 * 1.4 =134W
but i know this answer is wrong tho, the answer given is 629 W.

Welcome to PF.

I would suggest recalculating your frictional resistance. That looks a little light.

Thanks for the heads up, i didn't multiply with gravity as i should have - the friction force corrected is 750.5N

I think the parallel and net forces also play a role here in resisting friction. I can't seem to get the applied force down the inclined plane.

the parallel force along the incline is mg sin 23 = 497.8N

This should reduce the force acting down the incline ( F friction - f parallel) = 252.7 N
but friction is still more and it should not move.

how can i find the force needed to drag it down the incline at v= 1.4m/s? won't i need acceleration to find the net force?

There is no acceleration as it is at constant speed of 1.4 m/s.

So yes you do subtract the Sin23*130*9.8 from the frictional maximum need to push it along, because as you have figured it is the net force that HE supplies that is doing the work that they ask for.

I would note however that :

.8 * 130 * 9.8 * cos23 = 938.1785

omg, i got it! no net force, he just need to apply additional force of 449 N to overcome friction and move it at a constant speed. Multiply this force and velocity will give me the Power ! thanks :D

yup caught that too! it was messing up my calculations! thanks so much lowlypion! :D

solved!

## 1. How do you calculate power on an inclined plane?

In order to calculate power on an inclined plane, you will need to know the velocity of the object (v), the angle of the incline (θ), and the coefficient of friction (μ). The formula for calculating power on an inclined plane is P = (Fv)cosθ, where F is the force applied and v is the velocity.

## 2. What is the role of velocity in calculating power on an inclined plane?

Velocity (v) is an important factor in calculating power on an inclined plane because it represents the speed at which the object is moving along the incline. The greater the velocity, the more power is required to maintain that speed against the force of gravity and friction.

## 3. How does the coefficient of friction affect power on an inclined plane?

The coefficient of friction (μ) is a measure of the resistance between two surfaces in contact. On an inclined plane, this resistance plays a significant role in determining the amount of power required to overcome it. A higher coefficient of friction means more power is needed to move the object along the incline.

## 4. Can you calculate power on an inclined plane without knowing the coefficient of friction?

No, it is not possible to calculate power on an inclined plane without knowing the coefficient of friction. This value is essential in determining the amount of resistance that needs to be overcome, and without it, the calculation would be incomplete.

## 5. How does the angle of the incline affect power on an inclined plane?

The angle of the incline (θ) is another crucial factor in calculating power on an inclined plane. As the angle increases, the force of gravity pulling the object down the incline also increases, requiring more power to maintain the same velocity. Therefore, a steeper incline will require more power than a shallower one.

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