Pushing Crate up Frictionless Slope Find Work

In summary, a man is pushing a crate weighing 75 N up a frictionless slope at an angle of 30° with a force parallel to the slope. The crate is accelerating upward at a rate of 0.5m/s^2. The work done by the man while moving the crate a distance of 1.5m is 41J. This is calculated by resolving the weight of the crate into horizontal and vertical components, taking into account the acceleration and mass of the crate.
  • #1
Spartan Erik
31
0

Homework Statement



A man pushes a create which weighs 75 N upward along a frictionless slope that makes an angle of 30° with the horizontal. The force he exerts is parallel to the slope. If the speed of the crate increases at a rate of 0.5m/s^2, then the work done by the man while moving the crate a distance of 1.5m is:

-62J, +62J, -41J, +41J, none of the above

Homework Equations



W = Fdcos(theta)
W(done by gravity)= mgdcos(theta)

The Attempt at a Solution



I'm not sure exactly how to approach this problem as they supply an increasing acceleration of 0.5m/s^2. Otherwise I would just apply W = Fdcos(theta). I also know that gravity has an effect on this, but I'm not sure how to include it in this situation
 
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  • #2
Draw a free body diagram and resolve the weight of the box into components.

Weight pointing down the plane = mg sin theta
Weight normal to the plane (the normal force) = mg cos theta.

If the object is accelerating upward at .5m/s^2, what does that tell you about the net force on the object?
 
  • #3
Well the 75N is due to gravity, so the mass is 7.653kg

Breaking it into horizontal and vertical components:
mgsin(theta) = 7.653x9.8xsin(30) = 37.5
mgcos(theta) = 7.653x9.8xcos(30) = 64.951

If the object is accelerating upward at 0.5 m/s^2 and the mass is 7.653kg, then the force is 3.826N
 

1. How does friction affect the work done in pushing a crate up a frictionless slope?

Friction does not have any effect on the work done in pushing a crate up a frictionless slope. This is because there is no frictional force acting against the motion of the crate, so no work is required to overcome it.

2. What is the formula for calculating work done in pushing a crate up a frictionless slope?

The formula for calculating work done is W = mgh, where m is the mass of the crate, g is the acceleration due to gravity, and h is the height of the slope.

3. Can the work done in pushing a crate up a frictionless slope be negative?

No, the work done in pushing a crate up a frictionless slope cannot be negative. This is because work is defined as the product of force and displacement, and in this scenario, the force applied is always in the direction of displacement, resulting in a positive value for work.

4. How does the angle of the slope affect the work done in pushing a crate up a frictionless slope?

The angle of the slope has no effect on the work done in pushing a crate up a frictionless slope. As long as the slope is frictionless, the work done will only depend on the mass and height of the slope, not the angle.

5. Does the speed at which the crate is pushed up the frictionless slope affect the work done?

No, the speed at which the crate is pushed up the frictionless slope does not affect the work done. The work done only depends on the mass and height of the slope, not the speed of the crate.

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