# Law of Conservation of Energy

## Homework Statement

A 2.0 kg load has an initial velocity of 0.65m/s. If a frictional force acts to slow it down, how fast is it sliding down the inclined plane just before it reaches the ground? The coefficient of friction between the load and the inclined is 0.30.

Given/Known Data:
M: 2.0 kg
Velocity (Initial): 0.65m/s
Coefficient: 0.30

## Homework Equations

f=$\mu$Fn
f=$\mu$Wy
f=$\mu$mg(cos$\Theta$)

## The Attempt at a Solution

=(0.30)(2.0kg)(9.8m/s2)(cos 30)
= 0.30 (19.6N) 0.866
= 5.88N(.866)
= 5.09208N

Workagainstfriction
= 5.09208(2m)
= 10.18416Nm
= 10.18416J

ΔKE = 1/2mv^2 - 1/2mv^2 <--- This is the part where I screw up.

Any help will do. Thanks

## Answers and Replies

Erm, in your workings:
5.09208(2m) which I assume is force x distance, correct? Where did you get the value of distance from? Then I might be able to help.

This is the hypotenuse I got from the triangle formed by the height given which is 1m and the angle 0.30 degrees. So the 2m is where the load slides down.

From that, I'd assume constant acceleration. Find the decceleration due to the frictional force, and apply one of the constant acceleration equations, you should get the correct value, don't really know why you're using Work and Kinetic Energy.