How Fast Does a 100 kg Block Ascend a 5 m Slope?

AI Thread Summary
A 500 kg block is released from rest on a 20-degree slope, raising a 100 kg block while experiencing a coefficient of kinetic friction of 0.2. The user struggles to calculate the velocity of the 100 kg block after ascending 5 m, initially using an incorrect formula for acceleration. They are advised to draw free body diagrams for both masses and apply Newton's second law to derive the correct equations. The discussion emphasizes the importance of accurately accounting for both masses and the tension in the connecting string. The user is encouraged to take their time and seek further clarification on the concepts involved.
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Homework Statement



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


Homework Equations



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

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 :smile:

this is what i did so far and it really doesn't look right

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

= 264.63m/s (first thing that makes me think I am 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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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.
 
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
 

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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?
 
no i don't unfortunately

i don't 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 doesn't have a weight to raise up. Thats where my equation has gone wrong (amongst other things i guess)
 
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.
 
Ok i'll try and draw a free body diagram

Bare with me cos I am not that great at this
 
Take your time, but note that two (not one) FBDs are needed.
 
Ok here are my FBD's. I don't know if they are right tho unfortunately
 

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  • #10
I cannot see the FBDs "pending approval."
 
  • #11
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/second-Newton-law-exercises-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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