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## 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=[itex]\mu[/itex]F

_{n}

f=[itex]\mu[/itex]W

_{y}

f=[itex]\mu[/itex]mg(cos[itex]\Theta[/itex])

## The Attempt at a Solution

=(0.30)(2.0kg)(9.8m/s

^{2})(cos 30)

= 0.30 (19.6N) 0.866

= 5.88N(.866)

= 5.09208N

Work

_{againstfriction}

= 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