A 34.0 kg seal at an amusement park slides down a ramp into the pool below. The top of the ramp is 1.55 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.
(a) What is the work is done by kinetic friction?
(b) What is the coefficient of kinetic friction between the seal and the ramp?
I am having some trouble with this one. Any help would be much appreciated. Thanks!
2007-10-25 05:04:48 · 4 answers · asked by Tana 1 in Science & Mathematics ➔ Physics
Calculate the length of the ramp (1.55/cos(30)) and the gravitational acceleration parallel to the ramp (9.8*cos(30)). Then calculate the total acceleration acting on the seal to get it to 4.9 m/s at the end of the ramp (v² = 2ax). The difference between the two values is caused by friction. Calculate the force difference (F=ma) and you have the frictional force. This force acting through the length of the ramp is the work involved (W=Fx). To get the coefficient of friction, calculate the force of the seals body against the ramp (F=mg*cos(30)) and then the coeficient of friction is that value divided into the frictional force.
HTH
Doug
2007-10-25 05:33:54 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
It's been a while since I've done problems like this, but what about finding the energy at the beginning of the problem and the end. Potential energy at the beginning and kinetic energy for when the seal hits the water. The difference is the work done by friction. This will give you same units as work at least. Then, use W=force x distance with F=mass x acceleration x coefficient of friction. Solve for friction.
I don't know for sure if this is right, hopefully it helps.
2007-10-25 05:47:16 · answer #2 · answered by polyman2001us 1 · 0⤊ 0⤋
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2016-12-30 05:18:22 · answer #3 · answered by ? 3 · 0⤊ 0⤋
work is forceX distance
force= mass* accel of gravity= 34kg*9.81m/s/s=333.54N
work= 333.4*1.5 n-m= 500.31
This is a trick I learned from Aerospace
cof= (sped from gravity-speed form slide)/speed from gravity
distance=1/2 (9.81)t^2
1.5=1/2(9.81)t^2
t=.553 sec
velocity = 9.81 m/s/s*.553=5.424=
therefore cof = 0.1(untiles)
2007-10-25 05:20:54 · answer #4 · answered by eric l 6 · 0⤊ 0⤋