# Question regarding tension and acceleration

1. Nov 30, 2004

### Sanosuke Sagara

I have my question and my solution in my attachment.don't worry,just unzipped the file and no virus contains inside.

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2. Nov 30, 2004

### Sanosuke Sagara

Help,Somebody help me figure out whether my solution for the question is right or not.
Thanks for anyone who spend some time on this question.

3. Nov 30, 2004

### Sanosuke Sagara

Anybody please help me !! My solution for the question is inside the attachment.just unzipped it .

4. Nov 30, 2004

### Andrew Mason

Since there is no acceleration, initially, The force applied by the spring is equal to gravity:

F - (x + y)g = (x + y)a = 0 where F is the force of the spring.

F = (x+y)g

when y is cut loose, F remains the same, initially (F = kd)

F - xg = xa

So a = F/x - g

Since F = (x+y)g

a = ((x + y)g/x) - g

a = gy/x

Substituting values for x = .5 and y = .1:

a = 10 (.1)/.5) = 2 m/sec2

This makes sense, since the force that pulls the spring down from its equilibrium position with just x on the spring, is just the force of gravity on y.

AM

Last edited: Nov 30, 2004
5. Nov 30, 2004

### Staff: Mentor

I don't understand your solution. Start by identifying the forces acting on the top mass before the string is burned. Then find the net force on the top mass after the string is burned. Apply Newton's 2nd law.

Edit: This was meant for Sanosuke Sagara; but Andrew has solved it for you.

Last edited: Nov 30, 2004
6. Dec 1, 2004

### Sanosuke Sagara

Thanks for your help,Andrew Mason and mentor Doc Al.I finally have understand with the question.