Push a 22kg crate up a frictinless incline

[EmoryGirl]

Homework Statement

To push a 33 kg crate up a frictionless incline, angled at 15° to the horizontal, a worker exerts a force of 100.4 N, parallel to the incline. As the crate slides 1.50 m, (a)how much work is done on the crate by the worker's applied force? (b) How much work is done on the crate by the weight of the crate? (c) How much work is done on the crate by the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

Homework Equations

I can NOT solve (d)!

The Attempt at a Solution

I have solved a, b, and c:
(a) 1.51 x 10^2 J
(b) -1.26 x 10^2 J
(c) 0.00 J
(d) ?

 

[alphysicist]

Hi EmoryGirl,

Homework Statement

To push a 33 kg crate up a frictionless incline, angled at 15° to the horizontal, a worker exerts a force of 100.4 N, parallel to the incline. As the crate slides 1.50 m, (a)how much work is done on the crate by the worker's applied force? (b) How much work is done on the crate by the weight of the crate? (c) How much work is done on the crate by the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

Homework Equations

I can NOT solve (d)!

The Attempt at a Solution

I have solved a, b, and c:
(a) 1.51 x 10^2 J
(b) -1.26 x 10^2 J
(c) 0.00 J
(d) ?

The total work done on the crate is the work done by all the forces that are acting on the crate. What would that be?

 

[EmoryGirl]

Hi EmoryGirl,




The total work done on the crate is the work done by all the forces that are acting on the crate. What would that be?

The forces acting on the crate include the applied force and mg. I'm not sure how they work together in an equation...
I tried just W=Fd using 100.4N as the the F but I know that is not right

 

[alphysicist]

The forces acting on the crate include the applied force and mg. I'm not sure how they work together in an equation...
I tried just W=Fd using 100.4N as the the F but I know that is not right

It's not right because it is only the work from the applied force. Remember that you already have the work done from each individual force that is acting on the crate; what then is the total work done?

 

[EmoryGirl]

Okay I just solved it! I added the work done by the applied force to the work done by the weight of the crate. The answer is 25 J...I didn't realize that it was so simple!
Thanks for your help!

 

[alphysicist]

Okay I just solved it! I added the work done by the applied force to the work done by the weight of the crate. The answer is 25 J...I didn't realize that it was so simple!
Thanks for your help!

Glad to help, and welcome to PhysicsForums!