Calculating Work and Coefficient of Kinetic Friction for a Sliding Seal

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Homework Help Overview

The problem involves calculating the work done by kinetic friction and the coefficient of kinetic friction for a seal sliding down a ramp into a pool. The scenario includes a 43-kg seal, a ramp height of 1.75 m, and an incline angle of 35.0°.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss energy conservation principles, including initial and final total energy calculations. Some express frustration over the accuracy of their results and seek alternative methods or insights.

Discussion Status

There are multiple interpretations of the problem, with participants exploring different approaches to the calculations. Some have received feedback indicating that the problem may have been flawed, leading to a lack of consensus on the correct answers.

Contextual Notes

Participants note that the homework is submitted online, and while the due date has passed, they are still allowed to check their work. There is mention of a communication from the teacher regarding the problem being broken.

askthefool
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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)


~~
 
This is homework that's submitted online. The due date is already passed, and I missed the problem but that doesn'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 doesn'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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