Releasing a 4kg Weight: Can You Solve the Equation?

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The discussion revolves around solving a physics problem involving a 4kg weight released and the forces acting on it. Participants explore various methods to calculate acceleration, including using SUVAT equations and Newton's second law. The importance of considering tension in the string and the forces acting on both the 4kg and 6kg masses is emphasized. Confusion arises regarding the correct application of forces and the direction of tension, leading to a collaborative effort to clarify the equations. Ultimately, the conversation highlights the need for careful analysis of forces and their interactions in solving the problem.
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Homework Statement


physicsquestion.gif


The 4kg weight is released when the question begins


Homework Equations


kE = 1/2mv^2


The Attempt at a Solution


Had a go using SUVAT equations and came up with a velocity of around 22ms ready to put into the kE formula, then realized I hadn't done anything with the 4 seconds and can't figure out where it fits in.
 
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You need to start by calculating the acceleration of the masses.
 
I see 3 options

1. Use suvat
2. Final V - Initial V/Total time (I don't know any velocities)
3. Drop AS PhysicsI'm lost
 
I see a fourth option;

4. Newton's second law
 
Ok, now I have the mass, but not the acceleration or the force. Is this where I use gravity?
 
grinman said:
Ok, now I have the mass, but not the acceleration or the force. Is this where I use gravity?
Indeed, you need to examine the forces acting on each mass. I suggest you consider each mass as a separate entity and draw a separate FBD for each.
 
Ok, so I started with the 4kg weight and got a 40N force as gravity, or am I doing something wrong here
 
grinman said:
Ok, so I started with the 4kg weight and got a 40N force as gravity, or am I doing something wrong here
Nope, that looks good to me; but is weight the only force acting here?
 
Hmm, so I was thinking that the other force would be the 6kg weight pulling downwards at 60N, do I take the leftover 20N and put that into f=ma?
 
  • #10
grinman said:
Hmm, so I was thinking that the other force would be the 6kg weight pulling downwards at 60N, do I take the leftover 20N and put that into f=ma?
Whoa... slow down there. I said that we should consider each mass seperately, so what's the other force acting on the 40kg mass besides it's weight?
 
  • #11
Damn, I thought I was on to something there.

I really have no idea about the other force now, it's probably something really basic that I passed over
 
  • #12
grinman said:
Damn, I thought I was on to something there.

I really have no idea about the other force now, it's probably something really basic that I passed over
Okay, how about the tension in the string?
 
  • #13
Ah, right. I can't remember covering tension in my Physics class yet, though I seem to make a connection with a formula including change in length and original length, I'm not sure though.
 
  • #14
grinman said:
Ah, right. I can't remember covering tension in my Physics class yet, though I seem to make a connection with a formula including change in length and original length, I'm not sure though.
We're not really bothered about how the tension affects the string (we're assuming that the string inextensible), all we're really interested in is how it affects the mass. So, there are two forces acting on the 40kg mass, tension upwards and its weight downwards. This means we can write an expression using Newton's second law;

F_{net}=ma

T - mg = ma

T - 40g = 40a

Do you follow?
 
  • #15
So the tension is going to reduce the resultant force that causes the acceleration. I still can't see how I get the f=ma to work here though, T-40g=40a still leaves me with tension and acceleration unsolved :S What did I miss
 
  • #16
grinman said:
So the tension is going to reduce the resultant force that causes the acceleration. I still can't see how I get the f=ma to work here though, T-40g=40a still leaves me with tension and acceleration unsolved :S What did I miss
Well, what's say we have a look at the 60kg mass and do the same...?
 
  • #17
So we'd end up with T-60g=60a and T-40g=40a? I tried making a simultaneous equation here and ended up with T=240a and a=2 for the 40kg mass, is this good to go or did I miss the boat?

Also, shouldn't it be 4kg and 6kg instead of 40 and 60 (unless there was a reason that i missed)
 
  • #18
Anybody still online who can help me out?
 
  • #19
grinman said:
So we'd end up with T-60g=60a and T-40g=40a?
Careful here, you need to examine the direction of the forces. Is the tension really acting in the same direction that the 6kg mass is moving?
grinman said:
Also, shouldn't it be 4kg and 6kg instead of 40 and 60 (unless there was a reason that i missed)
Yes, sorry my bad, it should of course be 4kg and 6kg.
 
  • #20
So T+60=6a and T-40=4a, was I right about the simultaneous eq'n?
 
  • #21
grinman said:
So T+60=6a
Your closer but in which direction (relative to the motion) is the tension acting?
grinman said:
was I right about the simultaneous eq'n?
Yup, you were correct in your method.
 
  • #22
-t+60?
 
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