Post a New Question


posted by .

A 20N crate starting at rest slides down a rough 5.0m ramp, inclined at 25 degress with the horizontal. 20j of energy is lost due to friction. what will be the speed of the crate at the bottom of the incline?

i need some help these one i don't know where to start

For Further Reading

PHYSICS - drwls, Tuesday, October 23, 2007 at 7:41pm
Is your @0N (the weight, M g) supposed to be 20N? I will assume so. If not, you provide the correct value.

The loss in potential energy is
M g * 5 sin 25 meters = 42.3 J

PE loss = friction + kinetic energy gain

42.3 = 20 + KE increase = (1/2) M V^2
(1/2) M V^2 = 22.3 J
M = 20 N/9.8 = 2.04 kg
V^2 = 2*22.3/2.04 m^2/s62
solve for V

ok the first time i did it i got 3.2m/s and the second time i did to double check it i got 4.7m/s which is right?


    4.7 m/s is the correct answer.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question