How can power be the same on two inclined planes with different angles?

[alingy1]

I'm lost...

We have two inclined planes that take objects at the same heights. We need to lift a box. One of them has a higher angle. The one with the lower angle needs a longer displacement. That I know. The higher angle will need a bigger force.

In my textbook, it is written that since the work is the same in the two inclined planes (I know that the change in the height of the box is equal in the two planes), the power can be the same in the two inclined planes. We have to ignore friction. I'm in Grade 11.

So, I wonder how power can be the same? It is said that the time it takes to lift the box in both planes can be equal. How does that make sense? If you exerce a lower force on the box for a longer displacement, the time won't be equal to the plane with a higher angle! What's wrong here?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[PhanthomJay]

How is the problem worded exactly? If the object is pushed up either plane at constant velocity, the power can be the same if the time is the same, that is, it is pushed up the smaller angle incline at a faster speed than the larger angled incline, which is quite possible since you are applying less force. The problem is not clear as written, because the time could be more, less, or the same.

 

[darkxponent]

Time can be same. Did you verify it?
Power is same or different? You are just putting arguments. You should show some mathematical attempt.

Even your queation is not clear.

 

[alingy1]

How can the time applied be the same? You say that the speed could be faster in the smaller angle plane. But how can you have a higher speed with a specific force (that is, the force required to cancel the y-axis gravitational force)? If the box goes up at a higher speed, than the force applied to it increases, which increases the work done. In the two planes, the work done will not be equal anymore.

 

[voko]

When the velocity is constant, the net force is zero. The velocity can be arbitrary. So the time can be arbitrary.

 

[PhanthomJay]

How can the time applied be the same? You say that the speed could be faster in the smaller angle plane. But how can you have a higher speed with a specific force (that is, the force required to cancel the y-axis gravitational force)? If the box goes up at a higher speed, than the force applied to it increases, which increases the work done. In the two planes, the work done will not be equal anymore.

But why would you need to use a greater force at the higher constant speed?

 

[alingy1]

Well, you should apply a higher force to the box to give it the acceleration to reach that higher speed?

 

[PhanthomJay]

I see you have your thinking cap on in this unclear problem statement. If the object starts from rest on either incline, yes, you initially have to apply a force over a short period greater than the component of the gravity force down the incline to get it up to speed, but then the force is reduced when constant velocity is obtained. The time taken to reach the top on the small angle incline could be the same or more or less than the steeper incline. It doesn't have to be the same, but it could be.

 

[alingy1]

Ok, so, that's why the power can change and even be the same.
Then, the greater force than the x-axis of gravity to get the desired velocity is not considered in the calculation of the total work done, am I right?
Thanks a bunch for helping.

 

[voko]

One should also keep in mind that the problem is about just the work of lifting. If we simply leave the object alone as it reaches the top with a constant velocity, it won't just stop there, it will go higher than that. So we will have performed MORE work than required to lift the object to the top. To perform the exact amount of work required, we have to decelerate the object in such a way that it reaches the top with zero velocity. And decelerating is NEGATIVE work, and it will cancel out EXACTLY the work required for acceleration, no matter to what velocity. So all that is left is the work for MAINTAINING a constant velocity, i.e., keeping the net force at zero.