On slide 12 this question asks: a man pushes a 50 kg crate up a frictionless ramp with an inclined plane of 30 degrees and a length of 10m. How much work must be done to push the crate to the top?
The explanation says to use the workenergy theorem to calculate the change in potential energy as the work done to equal 2500 J. When I calculated it using Work = Fdcos(30) I got ~ 4300J what did I do wrong in my calculation??
Thank you!
Work and energy quiz

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Re: Work and energy quiz
I'm going to guess it is the trigonometry that is tripping you up.
A 30/60/90 right triangle, such as the crosssection of a wedgeshaped ramp, has sides 1/root3/2. If the ramp is 10 meters long, therefore the height is half that: 5 meters.
(Or in mathematical notation (sin30)*10=5)
So we arrive at mgh = 50*10* 5 = 2500 J
Let me know if this helps, or if the hangup was elsewhere!
A 30/60/90 right triangle, such as the crosssection of a wedgeshaped ramp, has sides 1/root3/2. If the ramp is 10 meters long, therefore the height is half that: 5 meters.
(Or in mathematical notation (sin30)*10=5)
So we arrive at mgh = 50*10* 5 = 2500 J
Let me know if this helps, or if the hangup was elsewhere!