Calculating Work and Coefficient of Kinetic Friction for a Sliding Seal

AI Thread Summary
A physics problem involving a 43-kg seal sliding down a ramp has generated frustration due to incorrect calculations of work done by kinetic friction and the coefficient of kinetic friction. The initial height of the ramp is 1.75 m, and the seal reaches the water at a speed of 4.40 m/s. The calculations for work done by friction resulted in -321.21 J, while the coefficient of kinetic friction was found to be 0.3050. Ultimately, it was revealed that the problem was flawed, and all students received credit despite the confusion. The discussion highlights the challenges of solving physics problems and the importance of verifying calculations.
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I've been doing this problem for the last 170 minutes. Literally, and I'm pissed.

A 43-kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.75 m higher than the surface of the water and the ramp is inclined at an angle of θ = 35.0° above the horizontal. If the seal reaches the water with a speed of 4.40 m/s, what is (a) the work done by kinetic friction and (b) the coefficient of kinetic friction between the seal and the ramp?

http://www.askthefool.com/Pic7.jpg

I've been doing W(non conservative) = 1/2 mv ^ 2 - mgh

h = 1.75/sin(35)

Don't know what's wrong.
 
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I'd like help ASAP. I'm going to bed soon, so the sooner the better. I appreciate it.
 
A 43-kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.75 m higher than the surface of the water and the ramp is inclined at an angle of θ = 35.0° above the horizontal. If the seal reaches the water with a speed of 4.40 m/s, what is (a) the work done by kinetic friction and (b) the coefficient of kinetic friction between the seal and the ramp?
{Initial Total Energy} = (1/2)*m*(v_i)^2 + m*g*h_i = 0 + (43)(9.8)(1.75) =
= (737.45 J)
{Final Total Energy} = (1/2)*m*(v_f)^2 + m*g*h_f = (1/2)*(43)*(4.40)^2 + 0 =
= (416.24 J)
{Delta Total Energy} = {Final Total Energy} - {Initial Total Energy} =
= (416.24 J) - (737.45 J) =
= (-321.21 J)
= {Work Performed By Friction} = {-K*m*g*cos(35 deg)}*{1.75/sin(35 deg)} =
= {-K*(43)*(9.8)*cos(35 deg)}*{1.75/sin(35 deg)} =
= K*(-1053.188)

A) {Work Performed By Friction} = (-321.21 J)
B) {Coeff of Kinetic Friction} = (0.3050)


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This is homework that's submitted online. The due date is already passed, and I missed the problem but that dosen't matter. It's only one point.

Howerver, we are still allowed to submit answer to check our work after the due date not for points, and i just tried the answers you provided and they are wrong. I've been getting those answers all night, trying a variety of ways.

Any other ideas?
 
I take that back. You/me are both right. The teacher emailed us saying that that problem was broken and that everyone got credit for it. Thanks for you're help.
 
askthefool said:
This is homework that's submitted online. The due date is already passed, and I missed the problem but that dosen't matter. It's only one point.

Howerver, we are still allowed to submit answer to check our work after the due date not for points, and i just tried the answers you provided and they are wrong. I've been getting those answers all night, trying a variety of ways.

Any other ideas?

I take that back. You/me are both right. The teacher emailed us saying that that problem was broken and that everyone got credit for it. Thanks for you're help.

haha, I was about to say that your teacher is wrong when i read your first reply cause xanthym is correct lol :approve:
 
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