What Went Wrong in Solving the Energy Equation Problem?

[Ut Prosim]

Homework Statement

The 10kg block has an initial velocity of 2m/s at the bottom of the ramp and slides up the ramp with a constant applied horizontal force F=75N as shown. Find the maximum height h that the block can reach on the ramp.

We're using cos(theta/%)=.8 & sin(theta/%)=.6

m=10kg | V₁=2m/s | F=75N | Y₂=? | Y₁=0 | coefficient of friction(#)= .4

Homework Equations

K₁ + U_g1+ U_e1+ W'_1-2= K₂ + U_g2 + U_e2

The Attempt at a Solution

K₁= .5m(V₁)² V₁=2 ==20
K₂= .5m(V₂)² V₂=0 ==0
U_g1= mgY₁ Y₁=0 ==0
U_g2= mgY₂ = 98Y₂

(elastics are 0)

W'_1-2
= W_F+W_f
= [-#*mgcos(%)*(y₂/sin(%))] + [Fcos(%)*(Y₂/sin(%))]
=[(-.4)(10)(9.8)(.8)*(Y₂/.6)] + [(75)(.8)*(Y₂/.6)]
=47.733Y₂Energy Equation:

20+0+0+47.733Y₂ = 0 + 98Y₂ + 0
Y₂=.398
BUT the answer is .249m.

If you could review my work, and help me find out where I went wrong, it would be greatly appreciated.

 

[KLoux]

Well... I don't get your answer, and I also don't get 0.249 m. The difference between my answer and yours comes from the energy equation and they way work is calculated. It looks like in your equation, the work done by friction helps and the work done by the external force hurts...

But like I said, I still don't get 0.249 m. It's always possible that I'm doing something wrong, too.

Also, it helps to use LaTeX for readability (it's hard to look at % and #... especially when you know it could be \theta and \mu). I might be nitpicking, but I also like to use the same variables that you're given in the problem. If it asks you to solve for h, why change it to Y_{2}?

-Kerry

 

[Ut Prosim]

The work done by friction hurts, hence it has negative mu . I just have them switched in the addition problem above.

I have no clue what LaTeX is, although I see when I try to copy and paste your mu, LaTeX Code: \\mu, it shows up as an odd code. I'm new to these forums and have not been introduced to this character code.

 

[LowlyPion]

I don't think you have enough terms. Also I think solving for the distance up the ramp is the way to go and then figure the h after you know how far up it goes.

To do this figure the net force on the block on the ramp:

Total-F = Applied-F*Cosθ - mg*sinθ - Friction-F

First determine what's happening with the first 2 forces.

75*.8 - 98*.6 = 1.2 N - friction

Now Friction is

μ*(75*sinθ + 98*cosθ) = .4(75*.6 + 98*.8) = 49.36

Net Force then = 1.2 - 49.36 = -48.16

From your force now you have the deceleration and can figure the distance by

V² = 2*a*x

Then just figure h from x.

 

[KLoux]

The work done by friction hurts, hence it has negative mu . I just have them switched in the addition problem above.

I have no clue what LaTeX is, although I see when I try to copy and paste your mu, LaTeX Code: \\mu, it shows up as an odd code. I'm new to these forums and have not been introduced to this character code.

Ah, yes, I believe you're right. I mis-read your post.

For LaTeX help, click on the \Sigma to the right of the array of buttons when you're postion.

LowlyPion found our mistake (I made the same one you did...): We didn't include the effect of the applied force F on the normal force when we calculated the friction force. If you do that, you should get the right answer (the same one you'd get with LowlyPion's method).

-Kerry