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

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To calculate the work done on a 50.0-kg cylinder pushed up a 3-m high, 6-m long ramp, the mass, height, and gravitational force are essential. The angle of elevation can be determined using the sine function, yielding 30°. The weight of the cylinder is calculated as 490 N, leading to a force of 245 N along the ramp. The work done is derived from the potential energy change, resulting in approximately 1470 J. The discussion confirms that friction is not a factor in this scenario.
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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 mgh. You also know that the work done by non-conservative forces is equal to the change in mechanical energy of the object. Therefore:
W = \Delta E_M = \Delta E_p = mgh = 50kg * 9.8\frac{m}{s^2} * 3m = 1470J
 

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