Incline, friction, and energy

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  • #1
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Homework Statement


An incline makes an angle of 29.9° with the horizontal. A 8.87 kg block is given a push up this incline and released. It starts at the bottom with initial speed 6.43 m/s, travels up the incline, stops, and slides back to the bottom at final speed 4.49 m/s. Using energy considerations, find:

a) the energy used to overcome friction traveling up and down the incline.
b) the distance the block traveled up along the incline before coming momentarily to rest.
c) The coefficient of kinetic friction between the block and the incline.


Homework Equations


Components (mgsinθ, mgcosθ)
W=Fs
P=Fv
KE=1/2*m(vf)^2 - 1/2*m(vo)^2


The Attempt at a Solution


I attached my FBD.

To find the energy, I would use the KE equation.

KE= 1/2*8.87*0 - 1/2*8.87*(6.43)^2 = (-183.4) J

So it took -183.4 J to climb up the incline which includes friction?

I'm not sure, but I guess we could do this:

P=Fv
-183.4 = (F)(6.43)
F = (-28.5N)

To find the distance (part b), W=Fs
(-183.4)=(-28.5)s
s=6.44m

Part (c), I'm confused..

Ff = μk*N
Ff+mgsinθ-F=μk*mgcosθ
 

Answers and Replies

  • #2
This problem is solved using 3 simple equations:

W[itex]\vec{F}[/itex]f=-F2[itex]\Delta[/itex]s=(.5)m(vf2-vi2) for a)

For b), use the previous equation (along with the energy you've calculated) and the following:

-(.5)mvi2+mgsinθ[itex]\Delta[/itex]s=-F[itex]\Delta[/itex]s

Finally, Ff=-μ.N with N=mgcosθ.

If you want to know the acceleration, use Newton's second law and:

a=-g(μcosθ+sinθ)
 
  • #3
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This problem is solved using 3 simple equations:

W[itex]\vec{F}[/itex]f=-F2[itex]\Delta[/itex]s=(.5)m(vf2-vi2) for a)

For b), use the previous equation (along with the energy you've calculated) and the following:

-(.5)mvi2+mgsinθ[itex]\Delta[/itex]s=-F[itex]\Delta[/itex]s

Finally, Ff=-μ.N with N=mgcosθ.

If you want to know the acceleration, use Newton's second law and:

a=-g(μcosθ+sinθ)

Hi Mathoholic,

It looks like I did part (a) incorrectly. Would it be that for initial velocity, I use 6.43 m/s, and for final velocity, I use 4.49? It would end up like this:

KE=1/2*m*(Vf^2-Vi^2)
KE=1/2*8.87*(4.49^2-6.43^2)
KE= (-93.95) J

(b) I don't exactly follow that formula, or why you're using it. When I plug in, I get:

-F=1/2*8.87*(6.43^2) + 8.87*9.81*sin(29.9)
-F=183.36 + 43.4 = 226.76
F= -226.76 N

For part (c), I don't quite follow how to solve it.
 
Last edited:
  • #4
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Fixed part (a). Noticed a huge mistake.

EDIT: Part (a) is still wrong..
 
  • #5
The part a) equation is correct, you're calculating how much work is done by the friction force along a path A→B→A (up and down the incline), in which [itex]\bar{AB}[/itex]=[itex]\bar{BA}[/itex]=Δs. In part b) you use the same equation but for half the way (A→B). That's why there's an additional term (mgsinθΔs), with Δh=sinθΔs.
You have two equations and two unknowns (F and Δs), hence, you can solve for both.
As for part c), you use Ff=-μN=-μ(mgcosθ) and substitude Ff by the value you've already computed before and you get μ.
 
  • #6
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The part a) equation is correct, you're calculating how much work is done by the friction force along a path A→B→A (up and down the incline), in which [itex]\bar{AB}[/itex]=[itex]\bar{BA}[/itex]=Δs. In part b) you use the same equation but for half the way (A→B). That's why there's an additional term (mgsinθΔs), with Δh=sinθΔs.
You have two equations and two unknowns (F and Δs), hence, you can solve for both.
As for part c), you use Ff=-μN=-μ(mgcosθ) and substitude Ff by the value you've already computed before and you get μ.

You say that my part (a) equation is correct, but what about the numerical value that I get at the end? I'm only asking because my online homework system tells me that it's wrong. I do..sort of understand why we use that equation though.
 
  • #7
You say that my part (a) equation is correct, but what about the numerical value that I get at the end? I'm only asking because my online homework system tells me that it's wrong. I do..sort of understand why we use that equation though.

Mind telling me what's your online homework system?

I'm absolutely sure that part a) is correct. Since the start position is the same as the ending position of the motion, you don't rely on potencial energy. Hence, you calculate the force done by friction using the variation of kinetic energy, which is what you've done.
 
  • #8
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Mind telling me what's your online homework system?

I'm absolutely sure that part a) is correct. Since the start position is the same as the ending position of the motion, you don't rely on potencial energy. Hence, you calculate the force done by friction using the variation of kinetic energy, which is what you've done.

webassign.net

That's what my school uses for assigning homework. I've entered -93.95. Should it be positive? I don't think so, right? What I mean is that maybe Vi and Vf should be:

KE=1/2*m*(-(Vf^2)-Vi^2)
KE=1/2*8.87*(-(4.49^2)-6.43^2)
KE= (-272.77) J

Vf is negative because it goes down to the left direction (assuming the push is up the incline toward the right direction, and we make the right side positive. So the push is positive (Vi is positive), then it comes back downward (Vf) in the negative. It obviously matters because the result for KE is much greater.
 
  • #9
webassign.net

That's what my school uses for assigning homework. I've entered -93.95. Should it be positive? I don't think so, right? What I mean is that maybe Vi and Vf should be:

KE=1/2*m*(-(Vf^2)-Vi^2)
KE=1/2*8.87*(-(4.49^2)-6.43^2)
KE= (-272.77) J

Vf is negative because it goes down to the left direction (assuming the push is up the incline toward the right direction, and we make the right side positive. So the push is positive (Vi is positive), then it comes back downward (Vf) in the negative. It obviously matters because the result for KE is much greater.

It can't be positive. The work done by friction is always negative. Kinetic energy is a scalar quantity, thus, it's independent of the direction of the motion. You cannot put (-(4.492)-6.432) because you're accounting for the direction of the vector velocity.
 

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