Work done by man pushing a crate up a ramp, includin done on

In summary, a man weighing 85 kg pushes a crate 4.00 m up a ramp at a 20 degree angle to the horizontal with a force of 500N, resulting in a negative magnitude. To determine the work done, the x and y components of the forces acting on the man are set up and the force applied is found to be parallel to the ramp. The work done is calculated to be 2000 J on the crate.
  • #1
SDTK
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Homework Statement


Calculate work done by an 85 kg man who pushes a crate 4.00 m up a ramp at an angle 20 degrees to horizontal.
He exerts a force of 500N on the crate parallel to the ramp, and moves at constant speed.
Include the work that he does on the crate and on his body to get up the ramp.

Homework Equations


I need to find the Force with which he moves himself up the ramp in other to determine the work he does on his body.
I set up x and y components of the forces acting on the man to find the Force, and have come up with a negative magnitude.

--Is setting the x and y components for the forces acting on the man the appropriate strategy?
-- do I include the force of the crate on the man?

The Attempt at a Solution


file attached
 

Attachments

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  • #2
Suppose for a moment that it was only the man on the ramp (no crate). What work is required to move him 4 meters up the ramp? What force is he working against? Hint: work-energy theorem.
 
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  • #3
gneill said:
Suppose for a moment that it was only the man on the ramp (no crate). What work is required to move him 4 meters up the ramp? What force is he working against? Hint: work-energy theorem.

thank you
 
  • #4
In your attached work you had this portion:
upload_2016-11-25_6-22-46.png


In the indicated equation what does the angle θ represent? Why did you set θ = 20° in the next line?
 
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  • #5
Should angle θ be "0"
because the force applied is parallel to the angle of the ramp?

W =F d cos(θ) = 500 kg 4 m cos(0) = 2000 J , work done on crate
 
  • #6
SDTK said:
Should angle θ be "0"
because the force applied is parallel to the angle of the ramp?

W =F d cos(θ) = 500 kg 4 m cos(0) = 2000 J , work done on crate
Yes. The force is applied in the same direction as the motion.
 

1. What is work done on an object?

Work done on an object is the amount of force applied to that object multiplied by the distance the object moves in the direction of the force. In other words, work done is a measure of the energy transferred to an object by a force.

2. How is work done by a man pushing a crate up a ramp calculated?

The work done by the man pushing a crate up a ramp is calculated by multiplying the force the man exerts on the crate by the distance the crate moves up the ramp in the direction of the force. This can be expressed as W = Fd, where W is the work done, F is the force, and d is the distance.

3. Is the work done on the crate the same as the work done by the man?

Yes, according to the law of conservation of energy, the work done on the crate by the man pushing it up the ramp is equal to the work done by the man. This means that the energy transferred to the crate is equal to the energy the man exerted in pushing it.

4. Does the angle of the ramp affect the amount of work done on the crate?

Yes, the angle of the ramp does affect the amount of work done on the crate. The steeper the ramp, the more force the man needs to exert to push the crate up, which means more work is done on the crate. The shallower the ramp, the less force and work are required.

5. How does friction affect the work done on the crate?

Friction opposes the motion of the crate and creates a force in the opposite direction of the force applied by the man. This means that more force and work are needed to overcome friction and push the crate up the ramp. As a result, friction decreases the amount of work done on the crate by the man.

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