Kinetic Friction of ride down a ramp

AI Thread Summary
A 28.0 kg seal slides down a 1.80 m high ramp inclined at 30.0°, reaching the water at 4.90 m/s. The initial gravitational potential energy is calculated as 493.92 J, while the final kinetic energy is 336.14 J, indicating a loss of energy due to friction. The work done by kinetic friction is determined to be -157.78 J after correcting for sign conventions. To find the coefficient of kinetic friction, the friction force must first be calculated using the relationship Wf = Ff * d, where Ff is dependent on the normal force. The discussion emphasizes the importance of correctly applying energy equations and sign conventions in physics problems.
MarineBio
Messages
15
Reaction score
0

Homework Statement



A 28.0 kg seal at an amusement park slides down a ramp into the pool below. The top of the ramp is 1.80 m higher than the surface of the water and the ramp is inclined at an angle of 30.0° above the horizontal. The seal reaches the water with a speed of 4.90 m/s.

What is the work done by kinetic friction?

Homework Equations



mgh

K = 1/2mv^2

The Attempt at a Solution



initial energy = mgh = 493.92

final energy = Kf = 1/2mv^2 = 336.14

If the initial was 493.92 and the final was 336.14 then I would assume there was a loss of 157.78, but that wasn't the right answer. Any help?
 
Physics news on Phys.org
I believe that you did this problem correctly. The only two things that I can think of are significant figures and minus sign conventions. I would call the work negative.
 
MarineBio said:

The Attempt at a Solution



initial energy = mgh = 493.92

my calculator does not give that, how did you get that?

But mgh= 1/2mv2 +Wf does look correctly applied.
 
turin said:
I believe that you did this problem correctly. The only two things that I can think of are significant figures and minus sign conventions. I would call the work negative.

Whoops, silly mistake. You're right, I just had to put a negative sign on the work of friction. Thanks.
 
Ok so the answer was -157.78 J and in order to find the coefficient of kinetic friction, uk, I thought it would be to simply divide by (mg) but I guess it isn't that simple in this scenario. How would I determine uk?
 
You would need to find the friction force first. You have that Wf = Ff * d and Ff = uk * FN
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top