Velocity of a block on a slope


by crazy31
Tags: block, slope, velocity
crazy31
crazy31 is offline
#1
Aug15-09, 08:39 AM
P: 5
1. The problem statement, all variables and given/known data

The block of mass 500 kg is released from rest at the top of the slope
It slides down a ramp of angle 20degrees
It raises a block weight of 100kg
The coefficient of kinetic friction between the 500 kg mass and the ramp is 0.2


2. Relevant equations

Determine the velocity of the 100 kg block when it has ascended 5 m

3. The attempt at a solution

Hi.

Im not very good with maths and physics and i'll freely admit that. I just need somehelp in prodding me in the right direction

this is what i did so far and it really doesnt look right

acceleration = 9.81 (sin20 - 0.2cos20)
(500-100)/1.51

= 264.63m/s (first thing that makes me think im doing the wrong thing)

then v squared = 0 + (2*264.63*5)

= 51m/s


Now even to someone as thick as me that looks completely wrong. So can someone help me out
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kuruman
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#2
Aug15-09, 08:48 AM
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How did you get the expression for the acceleration? It looks like the acceleration of a single mass sliding down the incline. It seems you did not take into account the 100 kg block.
crazy31
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#3
Aug15-09, 09:02 AM
P: 5
Here is a pic of the diagram in question


That acceleration part was something that i found in some notes for a question that i had done ages ago
Attached Thumbnails
Block weight.jpg  

kuruman
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#4
Aug15-09, 09:09 AM
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Velocity of a block on a slope


I cannot see the picture yet. Until I do, ask yourself whether the acceleration that you found in some old notes applies to the same situation as this one. Do you have a description of the old problem with a diagram that matches the picture that you posted?
crazy31
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#5
Aug15-09, 09:13 AM
P: 5
no i dont unfortunately

i dont even have a scanner to scan it into my computer and show you.


I did find the diagram that goes with the old equation and it doesnt have a weight to raise up. Thats where my equation has gone wrong (amongst other things i guess)
kuruman
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#6
Aug15-09, 09:21 AM
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That's OK. Eventually the picture will show up. Meanwhile,

1. Draw a free body diagram (FBD) for each of the masses.
2. Use the FBDs to write Newton's 2nd Law for each of the masses. This will give you two equations and two unknowns, the common acceleration of the masses and the tension in the string connecting the masses.
3. Eliminate the tension to find the common acceleration.
crazy31
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#7
Aug15-09, 09:26 AM
P: 5
Ok i'll try and draw a free body diagram

Bare with me cos im not that great at this
kuruman
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#8
Aug15-09, 09:38 AM
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Take your time, but note that two (not one) FBDs are needed.
crazy31
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#9
Aug15-09, 09:48 AM
P: 5
Ok here are my FBD's. I dont know if they are right tho unfortunately
Attached Thumbnails
FBD mass on slope.jpg   FBD2.jpg  
kuruman
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#10
Aug15-09, 10:28 AM
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I cannot see the FBDs "pending approval."
jeff1evesque
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#11
Aug15-09, 07:47 PM
P: 312
1. This is a good general picture of a free body diagram:
http://www.unm.edu/~caps/caps-handouts/free-body.html

2. I am going to guess what you're free body diagram should be (but am not entirely sure if I interpreted your problem correctly):
http://www.jfinternational.com/ph/se...ercises-2.html

3. I think if you read over those two links you will gain more an understanding- in particular, if you skim over the second link that should be sufficient I'd say.


Good luck,


JL


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