1 Answer | Add Yours
A ramp that is 6 m long is used to move a 2000 N crate onto a platform that is 2 m high. To do this a constant force of 1250 N has to be applied while the crate is moved.
Assuming the force is applied parallel in direction to the movement of the crate. The work done it moving the crate is equal to the product of the force applied and displacement of the crate. The work done is 1250*6 = 7500 J
The crate weighing 2000 N is moved by a distance of 2 m. This results in an increase in the potential energy of the crate of 2000*2 = 4000 J
As a work done of 7500 J increases the potential energy by only 4000 J, the efficiency of the ramp is 4000/7500 = 53.33 %
We’ve answered 319,199 questions. We can answer yours, too.Ask a question