# We did a lab in class to compare W and KE

• apaxapax
In summary, adding friction to the equation will cause the work done by the falling weight to be split between the kinetic energy imparted to the glider and the work done overcoming friction. The equations will show a lower velocity for the glider. The work done is the same.f

#### apaxapax

Hi All,

I am a real noob with physics...not my best topic. Trying to do the best I can. Just want to run my noob logic past some people much better at physics than me. We did a lab in class and I think I have the correct answers but just want some confirmation.

## Homework Statement

We did a lab in class to compare W and KE. We had a glider on a frictionless air rail (HORIZONTAL) and assumed no other friction forces acting on the glider. A string was attached to a hanging mass. We let the haning mass drop and we calculated W and KE for the glider and they were almost identical showing W=KE.

One question is if we add friction to the equation how will that effect the results. My thought is that the W and KE will increase.

The second question is if the rail was tilted forward or backwards how will that effect the results. My thought on that is if it was tilted backwards W and KE will show and increase and if tilted forward that the W and KE will show a decrease.

Do my thougts on for the answers make sense?

thanks for the help

## The Attempt at a Solution

Hi All,

I am a real noob with physics...not my best topic. Trying to do the best I can. Just want to run my noob logic past some people much better at physics than me. We did a lab in class and I think I have the correct answers but just want some confirmation.

## Homework Statement

We did a lab in class to compare W and KE. We had a glider on a frictionless air rail (HORIZONTAL) and assumed no other friction forces acting on the glider. A string was attached to a hanging mass. We let the haning mass drop and we calculated W and KE for the glider and they were almost identical showing W=KE.

One question is if we add friction to the equation how will that effect the results. My thought is that the W and KE will increase.

The second question is if the rail was tilted forward or backwards how will that effect the results. My thought on that is if it was tilted backwards W and KE will show and increase and if tilted forward that the W and KE will show a decrease.

Do my thougts on for the answers make sense?

thanks for the help

## The Attempt at a Solution

If you add friction to the equations, the work done by the falling weight (change in potential energy) will be split between the kinetic energy imparted to the glider and the work done overcoming friction. The equations will show a lower velocity for the glider. The work done is the same. You can only have as much work as the weight multiplied by the distance it falls.

For your second question, if the driving weight falls the same amount, the same amount of work is being done. In the case where the glider is pulled up a slope the work done by the falling weight now must be divided up between providing kinetic energy to the glider plus its change in potential energy because it has risen. That potential energy is the weight of the glider multiplied by the distance it rose.

In the case where the glider is going down a slope, its kinetic energy is changed by the work done by the falling weight plus the change in potential energy of the glider itself as calculated above.

For question one..I still don't understand why work will be the same. If you add friction work should increase? I attached a free body showing my work.

#### Attachments

• Free Body.pdf
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The work done by the falling weight on the system will be the same and it equals the change in the falling weight's potential energy. That is all that is available to go into the system. When there is no friction, all the potential energy (mgh) goes into kinetic energy of the airplane and the falling mass itself. When there is friction, the same energy (mgh) is divided up between the kinetic energies of both the plane, the falling weight, and the frictional work (F*distance).

I'll be away from the computer for the next few days.

Thanks for your help. Much appreciated