A rope is attached to a 50.0-kg crate to pull it up a frictionless incline with a 30 degree angle at a constant speed to a height of 3 meters.A: How long is the incline? B:Calculate the work done...

A rope is attached to a 50.0-kg crate to pull it up a frictionless incline with a 30 degree angle at a constant speed to a height of 3 meters.

A: How long is the incline?

B:Calculate the work done by NON-CONSERVATIVE FORCES.

C:What is the crate's potential energy when it reaches the top of the incline?

Asked on by newtonn

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justaguide | College Teacher | (Level 2) Distinguished Educator

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You have given that the rope is attached to a crate to pull it up a frictionless incline with a 30 degree angle at a constant speed to a height of 3 meters.

To find the length of the incline we use the relation for the sine of an angle. The crate rises to a height of 3m, so sin 30 = 3/L, where L is the length of the incline. sin 30 = .5 = 3 / L

=> L = 3/.5 = 6 m

The length of the incline is 6 m.

There is no work done by non-conservative forces as the incline is frictionless and the crate moves up the incline at a constant velocity.

When the crate reaches the top of the incline its height has increased by 3 m. This gives it a potential energy of m*g*h = 50*9.8*3 = 1470 J

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