Work dragging 50 kg mass up a 25° ramp with friction

[anyone1979]

[SOLVED] Plane facts, I

Is this right?

A ramp has an incline angle of 25 degress. How much work must you do if you are to drag a 50 kg mass a distance of 100m upward along the ramp? Assume that the coefficient of kinetic friction is uk = .35 and that you apply the force parallel to the plane.

uk mg cos(25) = (.35)(50)(9.8)cos(25) = 155N
Wfk = (155)(100)cos(180) = -15.5*10^3J

W = (100)(100)(1) = 1*10^4J

Wtot = 1*10^4 + 15.5*10^3 = 25.5*10^3J

 

[mgb_phys]

Not sure what the second equation (100)(100)(1) means
but you should have (friction force * linear distance) + (weight * vertical displacement)

 

[photonsquared]

Work

Assuming the coordinate system is established with the x-axis parallel with the ramp you would use the following:

W = F*d where

F = Fgravityx + Ff where

Fgravityx = mgsin(25)

Ff = u*N and

u = coefficient of friction
N = mgcos(25)

W = d(Fgravityx + Ff)

Hope this helps

 

[anyone1979]

Thank you for replying me.

The second statement is (Work = friction force * distance * cos 25 degrees)

When I get the answer, I found the total work done by adding the work done by friction and the work done by the force to get the total work done.

Is that right?

 

[mgb_phys]

Work is always force * distance.
The easiest way is to consider both the force to overcome friction * the 100m distance
Plus the vertical force (weight) * the vertical displacement ( 100m * sin(25) )

 

[CCdragonlord]

After a few rotations, the dragon has the same radius of rotation, but a shorter tail (period). Explain what effect this would have on the horizontal force acting on Jam:

 

[anyone1979]

Thank you all so much.
Correct me if I am wrong.

Isn't Wgravity = mgh
and h = d sin(25)?

That should make Wnet = Wgravity + Wf + Wfk
Which should give you negative for Wnet, but work cannot be negative?

 

[photonsquared]

The work is due to the Force of gravity plus the Force of friction. Not sure what your third force is in your equation. Your equation is correct for the force due to gravity.

 

[anyone1979]

Thank you for replying back so quick.
I am a little confused, the question asked to assume that you apply the force parallel to the plane. So I made the constant force applied equal to the distance (100N).
So, that third force was the work done by the force applied. Am I reading the question right, or should I remove the third force?