2 carts, a string, what is the work done?

  • Thread starter mujadeo
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  • #1
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Homework Statement



A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m, What is the work Ws done on the cart by the string?

Homework Equations


heres how it looks...

http://www.imagination3.com/LaunchP...7_211715035_342819690_usa&transcript=&_lscid=



The Attempt at a Solution


I know work is F * distance moved, and i can surmise that M1 moves the same distance as M2 (1m) cause theyre contrained by the string.
I know that force on M2 is 3 * 9.8 = 29.4N,
No friction so force on M1 (pulling horizontally) is also 29.4N
distance is 1 meter, so 29.4N * 1m = 29.4J
= wrong!

Please help!
]thanks
 
Last edited:

Answers and Replies

  • #2
learningphysics
Homework Helper
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Homework Statement



A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m, What is the work Ws done on the cart by the string?

Homework Equations





The Attempt at a Solution


I know work is F * distance moved, and i can surmise that M1 moves the same distance as M2 (1m) cause theyre contrained by the string.
I know that force on M2 is 3 * 9.8 = 29.4N,
No friction so force on M1 (pulling horizontally) is also 29.4N
distance is 1 meter, so 29.4N * 1m = 29.4J
= wrong!

Please help!
]thanks

Hint: The force acting on M1 is not the weight of M2... Draw the freebody diagram of M1... and also draw the freebody diagram of M2... label all the forces... What are the forces acting on M1... what are the forces acting on M2?
 
Last edited:
  • #3
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Yes i drew fbd for both, so i need to find the tension in string (on M1).

But I dont understand why force on m1 is not downwhard force (weight) on m2?

pully and surface are frictionless
strings are massless
i dont get it?
 
  • #4
learningphysics
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Yes i drew fbd for both, so i need to find the tension in string (on M1).

But I dont understand why force on m1 is not downwhard force (weight) on m2?

pully and surface are frictionless
strings are massless
i dont get it?

You would be right if the acceleration was 0... But that's not the case here. You have two equations with two unknowns... T (tension in the string) and a the acceleration (which is same for both carts).
 
  • #5
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why do i need the acceleration i just need the tension in the string dont i?
 
  • #6
Dick
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As in the last post I looked at you simply aren't writing the equations down. If you do, then what learningphysics is telling you will be much clearer.
 
  • #7
learningphysics
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why do i need the acceleration i just need the tension in the string dont i?

That's right, you don't need to calculate acceleration... But your two equations will have the two variables tension and acceleration. You only need to calculate the tension...

As Dick suggested, you should go ahead and write out the equations... it will make things much clearer... what do you get for your two equations?
 
Last edited:
  • #8
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M1
Fx = T = ma (no other x force cause no friction)
Fy = 0 (not moving vertically)
 
  • #9
learningphysics
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M1
Fx = T = ma (no other x force cause no friction)
Fy = 0 (not moving vertically)

That's right... and what equations do you have for M2?
 
  • #10
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Fx = 0 (not moving horiz)
Fy = T-W (mg) = ma
 
Last edited:
  • #11
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so my 2 unkonws are T and a, right?
 
  • #12
learningphysics
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Fx = 0 (not moving horiz)
Fy = T-W (mg)

Yes. I'd write it like this...

Fy = M2*a, therefore
T - M2*g = M2*a

and your equation for M1 from before is:

T = M1*a

So you have these 2 equations (the other 2 equations Fx=0 and Fy=0, don't help to solve the problem):

T - M2*g = M2*a
T = M1*a

And from here you can solve for T.
 
  • #14
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T-mg = ma (x)
T = ma (y)

is this ok so far?
 
  • #15
learningphysics
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T-mg = ma (x)
T = ma (y)

is this ok so far?

Yes, but be careful about the masses... write M1 and M2 instead of just m so that you aren't mixing up the masses...
 
  • #16
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OK so..

T-M1g = T

????

i combined equations but the Ts will just cancel
 
  • #17
learningphysics
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T-mg = ma (x)
T = ma (y)

is this ok so far?

It should be:

T - M2g = M2a
T = M1a
 
  • #18
learningphysics
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OK so..

T-M1g = T

????

i combined equations but the Ts will just cancel

You mixed up M1 and M2...
 
  • #19
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It should be:

T - M2g = M2a
T = M1a


M2a + M2g = M1a
M1a - M2a = M2g
a(M1-M2) = M2g

a= M2g / (M1-M2)

a = 9.8m/s/s

F=ma
58.8N = 6*9.8 (of M1)

W=Fx

58.8N * 2m = 117.6J

doesnt add up though????
 
  • #20
learningphysics
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M2a + M2g = M1a
M1a - M2a = M2g
a(M1-M2) = M2g

a= M2g / (M1-M2)

a = 9.8m/s/s

F=ma
58.8N = 6*9.8 (of M1)

W=Fx

58.8N * 2m = 117.6J

doesnt add up though????

Your original description says 1m, so it should be 58.8N * 1m = 58.8J. Is the answer 58.8J?
 
  • #21
103
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your right --thanks
but 58.8J is not right either.
??
 
  • #22
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M1 = 6kg
M2 = 3kg
h = 1m
 
  • #23
learningphysics
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I'm sorry mujadeo, I made a mistake... That second equation should be:

T - M2g = -M2a (taking up as positive and down as negative)...

So

T - M2g = -M2a
T = M1a

-M2a + M2g = M1a
M1a + M2a = M2g
a(M1+M2) = M2g

a= M2g / (M1+M2)

a=3.27...

so T = M1*a = 6*3.27= 19.6N

So work = 19.6N*1m = 19.6J... hope it's right.
 
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  • #24
103
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ah. add the masses.
yeah 19.6J is right, and i do undersant the whole process now

thanks alot!
 
  • #25
learningphysics
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ah. add the masses.
yeah 19.6J is right, and i do undersant the whole process now

thanks alot!

You're welcome. :smile: Do you see why I had to switch the sign? Since we labeled 'a' as being the acceleration of M1 to the right, we have to label 'a' as being the downward acceleration of M2... for consistency... then taking up as positive and down as negative... the acceleration of M2 is '-a'...
 

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