Quick question about work and forces(incline plane)

[vande060]

Homework Statement

A block of mass is pushed up an incline
of width d and height h with coefficient of friction μk
up to the top of the incline by a constant applied force of
magnitude F parallel to the ground. The block begins
at rest.

Homework Equations

f=ma

w = f . d

The Attempt at a Solution

if i choose a coordinate system with an x-axis Parrnell to the incline plane, then would it go like this:

normal force work= 0

applied force work = Fd/cos(60) <-- not sure about this one, my prof just wrote Fd in class, but i didnt understand why

work by fric = -ukmgd

work done by gravity = -mgh

is this right?

 

[PhanthomJay]

Homework Statement

A block of mass is pushed up an incline
of width d and height h with coefficient of friction μk
up to the top of the incline by a constant applied force of
magnitude F parallel to the ground. The block begins
at rest.

What is the question?

Homework Equations

f=ma

w = f . d

The Attempt at a Solution

if i choose a coordinate system with an x-axis Parrnell to the incline plane, then would it go like this:

normal force work= 0

this is correct

applied force work = Fd/cos(60) <-- not sure about this one, my prof just wrote Fd in class, but i didnt understand why

where did the 60 come from? And work is not fd/cos theta. Work is force times displacement times the cos of the angle between the 2. And if d represents the base width of the incline, the distance along the incline is from heel to toe is greater than d.

work by fric = -ukmgd

if you are calling d the distance traveled along the incline, then yes, correct.

work done by gravity = -mgh

yes, correct, that is the work done by gravity from the bottom to top of the incline

is this right?

Please post the question exactly as written and perhaps as illustrated.

 

[vande060]

the question is just to find the work done by each of the forces, just really to find the expression for each individual force. i modified the applied force (F), because it is not parallel to my chosen coordinate system, it is parallel to the ground.(picture on its way)

http://s861.photobucket.com/albums/ab174/alkaline262/?action=view&current=inclineplanejpg.jpg

i actually know think work done by Force applied = Fsin(theta)d

:D

 

[PhanthomJay]

the question is just to find the work done by each of the forces, just really to find the expression for each individual force. i modified the applied force (F), because it is not parallel to my chosen coordinate system, it is parallel to the ground.(picture on its way)

http://s861.photobucket.com/albums/ab174/alkaline262/?action=view&current=inclineplanejpg.jpg

i actually know think work done by Force applied = Fsin(theta)d

:D

This is not correct. Firstly, you seem to have broken up the applied force into its components parallel and perpendicular to the incline.. Your trig is wrong. The component of the applied force along the incline is F(cos theta).

Secondly, the wording indicates or seems to indicate that 'd' is the length of the base of the triangle, and not the length of the incline. So work done by the applied force F is, by definition, F(L) cos theta, where L is the length of the incline; and since from trig L =d/cos theta, the equation for work simplifies to W =Fd, which is actually an alternate definition for work: W = force times distance traveled in the direction of the force, which is positive if the displacement is in the same direction of the force, and negative if the didsplacement is in the opposite direction of the force.

 

[rwisz]

Your attempt at a solution is mostly correct. The only thing to note is that the angle between the applied force and the incline plane is not necessarily 60 degrees. It depends on the orientation of your coordinate system. If you choose the x-axis to be parallel to the incline, then the angle would be 90 degrees. If you choose the x-axis to be parallel to the ground, then the angle would be 0 degrees.

Other than that, your equations are correct. However, the work done by the applied force should be Fd, not Fd/cos(60). This is because the applied force is parallel to the displacement, so the angle between them is 0 degrees. Therefore, the cosine of the angle would be 1, and you can simply write the work as Fd.

Overall, your understanding of the concepts and equations is good. Keep up the good work!