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

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Homework Help Overview

The problem involves a 500 kg block sliding down a ramp at an angle of 20 degrees, which raises a 100 kg block. The coefficient of kinetic friction between the 500 kg mass and the ramp is 0.2. The goal is to determine the velocity of the 100 kg block after it has ascended 5 meters.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration using a formula found in previous notes but questions the validity of their approach given the context of the problem. Some participants question the appropriateness of the acceleration expression used, noting that it seems to apply to a single mass rather than the system involving both blocks.

Discussion Status

Participants are actively discussing the need for free body diagrams (FBDs) for both masses to apply Newton's second law correctly. There is an emphasis on drawing accurate diagrams and understanding the relationships between the forces acting on both blocks. Guidance has been offered on how to set up the equations based on the FBDs.

Contextual Notes

The original poster acknowledges a lack of relevant diagrams and admits to uncertainty about their calculations. There is a mention of missing information regarding the old problem that informed their initial approach.

crazy31
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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
 

Attachments

  • Block weight.jpg
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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
 

Attachments

  • FBD mass on slope.jpg
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  • FBD2.jpg
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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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