Push a 22kg crate up a frictinless incline

In summary, the conversation discusses a problem where a worker pushes a 33 kg crate up a frictionless incline at an angle of 15° to the horizontal. The worker exerts a force of 100.4 N parallel to the incline, causing the crate to slide 1.50 m. The question asks for the work done on the crate by the worker's applied force, the work done by the weight of the crate, the work done by the normal force exerted by the incline on the crate, and the total work done on the crate. The conversation concludes by finding the total work done on the crate to be 25 J by adding the work done by the applied force and the work done by the weight
  • #1
EmoryGirl
12
0

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) ?
 
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  • #2
Hi EmoryGirl,

EmoryGirl said:

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?
 
  • #3
alphysicist said:
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
 
  • #4
EmoryGirl said:
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?
 
  • #5
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!
 
  • #6
EmoryGirl said:
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!
 

1. How do you calculate the force needed to push a 22kg crate up a frictionless incline?

To calculate the force needed to push a 22kg crate up a frictionless incline, you can use the formula F = mgsinθ, where F is the force, m is the mass (22kg in this case), g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the incline.

2. What is the effect of the incline angle on the force needed to push the crate?

The force needed to push the crate will increase as the incline angle increases. This is because the component of the force acting against gravity (mg) will also increase, making the net force needed to push the crate up the incline greater.

3. Does the weight of the crate affect the force needed to push it up the incline?

Yes, the weight (or mass) of the crate directly affects the force needed to push it up the incline. The greater the weight of the crate, the greater the force needed to overcome gravity and push it up the incline.

4. Is the force needed to push a 22kg crate up a frictionless incline different from pushing it on a horizontal surface?

Yes, the force needed will be different because on a horizontal surface, the only force acting against the crate is its weight (mg). However, on an incline, the weight will have a component acting against the direction of motion, requiring a greater force to push the crate up.

5. How does the lack of friction affect the force needed to push the crate up the incline?

The lack of friction will not affect the force needed to push the crate up the incline, as friction is not a factor in this scenario. However, it is important to note that in real-world situations, friction will always be present and will require additional force to overcome.

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