Work Done on 50.0-kg Cylinder Up 3-m High, 6-m Ramp

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Homework Help Overview

The discussion revolves around calculating the work done on a 50.0-kg cylinder being pushed up a ramp that is 3 meters high and 6 meters long. The problem involves concepts from mechanics, specifically work, energy, and forces acting on an object on an inclined plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessary parameters to determine work done, including mass, friction, ramp dimensions, and angle of elevation. There is an exploration of trigonometric relationships to find the angle of the ramp and calculations involving gravitational force and potential energy.

Discussion Status

Some participants have provided calculations and interpretations of the work done, with one confirming the correctness of another's approach. There is mention of using potential energy in the absence of friction, and while some calculations have been shared, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants note that friction is not a factor in this scenario, and there is a focus on understanding the relationship between work and potential energy in the context of the ramp setup.

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Test tomorrow an I am sure somthing like this will be on it. What is needed to find out the work done upon a 50.0-kg cylinder that is being pushed up a 3-m high, 6-m long (hypotonuse) ramp?
 
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The mass of whatever's being pushed, the coefficient of friction between whatever's being pushed and the ramp, and any two of the following: the length of the ramp, the height of the ramp, the hypotenuse of the ramp, the angle of elevation of the ramp.

cookiemonster
 
Mind checkin this?

Is this correct? :

Sin^-1(3/6) = 30°

50.0 x 9.8 = 490-N

Sin30(490) = 245-N

W= 245 x 6 = 1476-J

Oh and there's no friction needed.
 
Last edited:
Ah... Looks right to me.

cookiemonster
 
If there is no friction, you can just use the potential energy of the object. You know that at the bottom of the ramp, its potential energy is zero. You also know that at the top of hte ramp, its potential energy is [itex]mgh[/itex]. You also know that the work done by non-conservative forces is equal to the change in mechanical energy of the object. Therefore:
[tex]W = \Delta E_M = \Delta E_p = mgh = 50kg * 9.8\frac{m}{s^2} * 3m = 1470J[/tex]
 

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