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?
1 Answer
justaguide | College Teacher | (Level 2) Distinguished Educator
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