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

  • Thread starter Thread starter chaotixmonjuish
  • Start date Start date
  • Tags Tags
    Ball Block
Click For Summary
A block slides along a curved track with friction on the flat section, and the problem involves calculating the maximum height reached on the opposite side after being released from a height of 82 cm. The discussion highlights the importance of using energy conservation principles, where potential energy at the start is converted into kinetic energy and then affected by friction on the flat part. Participants discuss calculating the work done by friction and how it impacts the final height. One user initially miscalculated the frictional force, leading to an incorrect height, but later corrected it after thorough analysis. The conversation emphasizes the application of physics concepts like potential and kinetic energy, as well as the role of friction in energy loss.
chaotixmonjuish
Messages
284
Reaction score
0
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
 
Last edited by a moderator:
Physics news on Phys.org
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
 
Last edited:
chaotixmonjuish said:
Won't this just calculate a velocity.

No. write out the equation.
 
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)
 
  • #10
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?
 
  • #11
.667
 
  • #12
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.
 
  • #13
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?
 
  • #14
img:http://i242.photobucket.com/albums/ff106/jtdla/prob21a.gif
 
Last edited by a moderator:
  • #15
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.
 
  • #16
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!
 
  • #17
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!
 

Similar threads

Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
7K
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
17K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
3K