What is the maximum height reached by a block on a curved track with friction?

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A 0.30-kg block slides along a small track with elevated ends and a flat central part. The flat part has a length L = 1.55 m. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is 0.106. The block is released from rest from a height h = 82 cm on the left curved portion of the track. Calculate the maximum height reached by the block on the right curved portion of the track.

img: http://i242.photobucket.com/albums/ff106/jtdla/prob09.gif

The only thing I have figured out is the Potential Energy at the top, then the KE at the bottom.

.3*9.8*.82=1/2*.3*x^2, that will yield a velocity. Outside of that, would I just treat the other part like a FBD
 
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Work done by friction = final energy - initial energy

final energy = Work done by friction + initial energy
 
Would I just calculate friction and then multiply it by the length of the bowl?

N=mg
muN=frictional force
muN*length of bowl
 
chaotixmonjuish said:
Would I just calculate friction and then multiply it by the length of the bowl?

N=mg
muN=frictional force
muN*length of bowl

yes. remember that the frictional work is negative... so work by friction is -muN*length of bowl.
 
by initial energy, do you mean potential energy
 
chaotixmonjuish said:
by initial energy, do you mean potential energy

yes. both initial and final energies are just potential energy...
 
Won't this just calculate a velocity.

I got 3.58 m/s

1/2*.3*x^2=-.31164*1.5+.3*9.8*.82
 
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Oh!

So would it be the potential energy of both sides of the bowl, except on one side i would have an unknown height

mgh (left)-work of friction=mgh(right)
 
chaotixmonjuish said:
Oh!

So would it be the potential energy of both sides of the bowl, except on one side i would have an unknown height

mgh (left)-work of friction=mgh(right)

yeah. what answer did you get?
 
A skier (m=59.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 7.70 m and D = 12.90 m, find H.

Could I solve this problem in a similar manner?

At the point where the skier is about to go off the ramp, is that both a PE and a KE.
 
chaotixmonjuish said:
A skier (m=59.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 7.70 m and D = 12.90 m, find H.

Could I solve this problem in a similar manner?

At the point where the skier is about to go off the ramp, is that both a PE and a KE.

I don't understand the question... what are h, D and H? can you describe or post the picture?
 
img:http://i242.photobucket.com/albums/ff106/jtdla/prob21a.gif
 
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chaotixmonjuish said:
A skier (m=59.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 7.70 m and D = 12.90 m, find H.

Could I solve this problem in a similar manner?

At the point where the skier is about to go off the ramp, is that both a PE and a KE.

Use kinematics to get the velocity at h when it goes off the ramp... then use energy conservation to get the height H.
 
My question is similar:

A 0.40-kg block slides along a small track with elevated ends and a flat central part. The flat part has a length L = 1.41 m. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is 0.145. The block is released from rest from a height h = 52 cm on the left curved portion of the track. Calculate the maximum height reached by the block on the right curved portion of the track.

I did:
Energy initial + work done by friction = Energy final
mgh + F*x = mgh

(mgh+F*x)
----------- = h
(mg)

[((0.4)(9.8)(0.52))+((0.145*-9.8)*(0.52))]
----------------------------------------- = h = 0.008875m
((0.4)(9.8))

But the real answer is: 0.31555m
Can anyone point out my mistake?
Thank you for the assistance!
 
I found my mistake... I used 9.8 as my force of friction. After 3 hours of doing this problems different ways... I finally found it! Thanks!