Work Done By a Gravitational Force

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SUMMARY

The work done by gravitational force on an object moving up a ramp is calculated using the formula -mgd, where m is mass, g is acceleration due to gravity, and d is the distance along the ramp. The discussion clarifies that the work done by gravity is equivalent to the change in gravitational potential energy, represented as -mgh, which can also be expressed as -mgdsin(theta) when considering the ramp's angle. The confusion arose from the distinction between horizontal and vertical components of gravitational force.

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xdarkelf714x
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Hi, I'm having trouble understanding this concept. A refrigerator is on a ramp of length d and it is being pushed up the ramp. What is the work done by the gravitational force? In the book it says mgdcos180 which is -mgd. I thought that the horizontal component of gravity was mgsin(theta) so the work done would by (mgsin(theta))(d)(cos180).

Also, is work done by the gravitational force difference than gravitational potential energy?

Thanks in advance.
 
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xdarkelf714x said:
Hi, I'm having trouble understanding this concept. A refrigerator is on a ramp of length d and it is being pushed up the ramp. What is the work done by the gravitational force? In the book it says mgdcos180 which is -mgd. I thought that the horizontal component of gravity was mgsin(theta) so the work done would by (mgsin(theta))(d)(cos180).
Assuming that d is the distance along the ramp and not the height, then you're correct. In terms of height (h), the work done by gravity is just F*S = -mgh, which is equivalent to -mgdsin(theta).

Also, is work done by the gravitational force difference than gravitational potential energy?
Same thing (just opposite sign).
 
Oh, I see it now. Thank you for clearing that up for me.
 

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