Posted by magic 8 ball on .
A @0N 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
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