Conservation of Energy, PE & KE, block on incline with KE & friction?

[nchin]

A 4 kg bundle starts up a 30° incline with 128 J of kinetic energy. How far will it slide up the plane if the coefficient of friction is 0.30?

So the solution to the problem is 128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30)...x = 4.29m

I don't understand the solution.

so 128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30) is

KE = mgh (sin theta) + Friction (mgh)(cos theta)

I thought it would be KE = PE + Friction, why is mgh added twice?

 

[lep11]

KE=PE-W where W is work done by friction.
KE=mgh-Friction*distance
Friction=normal force*coefficient of friction, now normal force is mgh*cos30

 

[frogjg2003]

The coefficient of friction is just a unitless factor that tells you how much the (maximum) force of friction compared to the normal force. What is the normal force?

 

[NoPoke]

It really is KE = PE + Losses to friction

have a closer look at the connection between the height above the starting point and how far along the slope the block travels

 

[nchin]

KE=PE-W where W is work done by friction.
KE=mgh-Friction*distance
Friction=normal force*coefficient of friction, now normal force is mgh*cos30

Isn't friction just

Coefficient x mg

Without height?

 

[frogjg2003]

For friction, F=μF_n. F_n isn't mg. Second, x isn't h. It's the distance the block slides along the surface. x sin(theta) is h. Third, x cos(theta) is also not h.
What you are calculating is not the force, but the work, which is force times distance. The force has an mg term in it, but the work has an mgh term.

 

[lep11]

Isn't friction just

Coefficient x mg

Without height?

Yes, it was a typo.

 

[lep11]

For friction, F=μF_n. F_n isn't mg. Second, x isn't h. It's the distance the block slides along the surface. x sin(theta) is h. Third, x cos(theta) is also not h.
What you are calculating is not the force, but the work, which is force times distance. The force has an mg term in it, but the work has an mgh term.

By the way you have a little typo too. F_μ=μF_n, now in y-direction ƩF=0⇔G_y=F_n →F_n=mg*cos30 So F_μ=μ*mg*cos30 Do you agree with me now?

 

[nchin]

By the way you have a little typo too. F_μ=μF_n, now in y-direction ƩF=0⇔G_y=F_n →F_n=mg*cos30 So F_μ=μ*mg*cos30 Do you agree with me now?

Ok that makes sense but why is the solution to the problem is:

128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30)

Which is
KE = mgh (sin theta) + Friction (mgh)(cos theta)

Shouldn't it just be
KE = (mgh) sin theta + coefficient of friction(mg) (cos theta)?

But that gives the wrong answer

But it would make more sense because friction = mu x mg

 

[frogjg2003]

By the way you have a little typo too. F_μ=μF_n, now in y-direction ƩF=0⇔G_y=F_n →F_n=mg*cos30 So F_μ=μ*mg*cos30 Do you agree with me now?

Were you talking about the part that looks like F_n*F_n? That's not a typo, it's two separate sentences.

 

[lep11]

Were you talking about the part that looks like F_n*F_n? That's not a typo, it's two separate sentences.

I am sorry, it really seemed to me Fn*Fn

 

[frogjg2003]

[KE] = (mgh) sin theta + coefficient of friction(mg) (cos theta)

Look at the units.

[KE]=J=kg*m²/s²

[m]=kg
[g]=m/s²
[h]=m
[sin(θ)]=unitless
[mgh sin(θ)]=kg*m²/s²

[μ]=unitless
[m]=kg
[g]=m/s/s
[cos(θ)]=unitless
[μmg cos(θ)]= kg*m/s²

Without going into any physics, your answer is wrong because the units don't agree.


Let's start from the top.

1. What are the FORCES acting on the bundle? Write these in terms of their components parallel and perpendicular with the slope.
2. What is the net force? It should be parallel to and down the slope.
3. Use $W=\int\vec{F}\cdot d\vec{x}$ to calculate the WORKas a function of the distance the bundle travels. Remember, $d\vec{x}$ is in the direction parallel to and up the slope, so your answer should be negative.
4. Solve KE+W=0 for x, the distance up the slope. This is NOT the distance in the horizontal direction, nor is it the height.