How Does Friction Affect Velocity on an Inclined Plane?

[andorei]

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=$\mu$F_n
f=$\mu$W_y
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

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

 

[TaxOnFear]

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.

 

[andorei]

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.

 

[TaxOnFear]

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.

 

[asteriatic]

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred from one form to another. In this problem, the initial kinetic energy of the load is being converted into work against friction, which is then being converted into potential energy as the load slides down the inclined plane. Therefore, we can use the equation for conservation of energy to solve for the final velocity of the load just before it reaches the ground.

Initial kinetic energy = Final potential energy
1/2mv^2 = mgh
Where m is the mass of the load, v is the final velocity, g is the acceleration due to gravity (9.8m/s^2), and h is the height of the inclined plane.

Substituting the given values, we get:
1/2(2.0kg)v^2 = (2.0kg)(9.8m/s^2)(2m)
Simplifying, we get:
v^2 = 19.6m^2/s^2
Taking the square root of both sides, we get:
v = 4.43m/s

Therefore, the load will be sliding down the inclined plane with a final velocity of 4.43m/s just before it reaches the ground.