Push a crate up a frictionless incline

In summary, the question is asking for the total work done on a 29 kg crate being pushed up a frictionless incline at a 39° angle by a force of 321.9 N parallel to the incline. The work done by the worker's force can be calculated using the equation W = Fdcos(theta), where F is the force applied, d is the displacement, and theta is the angle between the force and displacement. The work done by the weight of the crate and the normal force exerted by the incline can also be calculated using this equation. The total work done on the crate is the sum of these three values.
  • #1
almondb
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Homework Statement



To push a 29 kg crate up a frictionless incline, angled at 39° to the horizontal, a worker exerts a force of 321.9 N, parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by the worker's applied force?

How much work is done on the crate by the weight of the crate?

How much work is done on the crate by the normal force exerted by the incline on the crate?

What is the total work done on the crate?



I really don't know where to start on this. I tried breaking the gravity force into vectors but I'm not sure it that is right. Any suggestions on how to go about this?
 
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  • #2
Homework Equations W = Fdcos(theta)The Attempt at a Solution I'm not sure how to calculate the work done on the crate by the worker's applied force, the weight of the crate, and the normal force exerted by the incline.
 

1. What is the concept behind pushing a crate up a frictionless incline?

The concept behind pushing a crate up a frictionless incline is to overcome the effects of gravitational force and move the crate from a lower position to a higher position on the incline. This requires exerting a force on the crate in the direction of the incline.

2. What is the role of friction in pushing a crate up a frictionless incline?

In this scenario, friction does not play a role as the incline is assumed to be completely frictionless. This means that there is no resistance to the movement of the crate, making it easier to push up the incline.

3. How does the angle of the incline affect the amount of force needed to push the crate?

The steeper the incline, the more force is required to push the crate up. This is because the component of the crate's weight acting parallel to the incline increases as the angle increases, making it more difficult to overcome.

4. Is it possible to push a crate up a frictionless incline without applying any force?

No, it is not possible to push a crate up a frictionless incline without applying any force. This is because the crate would remain stationary due to the effects of gravity unless a force is exerted on it in the direction of the incline.

5. How does the mass of the crate affect the amount of force needed to push it up a frictionless incline?

The mass of the crate does not affect the amount of force needed to push it up a frictionless incline. This is because in this scenario, the weight of the crate does not contribute to the force needed to push it up the incline, as there is no friction to overcome.

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