# Homework Help: A tension based problem

1. Oct 8, 2007

### ideefixem

Greetings everyone.
This is my first post and I'm currently enrolled in a general physics I class. We are studying Newtons Second at the moment.

1. The problem statement, all variables and given/known data
A 50kg box hangs from a rope. What is the tension in the rope if:
a- The box moves up a steady 5.0 m/s
b- The box has vy= 5.0 m/s and is speeding up at 5.0 m/s^2

2. Relevant equations
a->= F->net/m
(Fnet)x=max
(Fnet)y=may

3. The attempt at a solution
I was out last Friday when this topic was discussed. The above question is one of the few review/summary questions available for extra practice. I don't want to fall behind so I'm just want to catch up for class tomorrow.

Any hints or ideas as to where to start would be much appreciated!

Thanks

2. Oct 8, 2007

### ideefixem

Alright,
Could you tell me if I am heading in the right direction here?

B)
m=50kg
v=5.0m/s
a=0m/s^2

f=ma=m(g+x)
f=50kg(9.8m/s^2+0)
tension=490N

C)
f=ma=m(g+x)
f=50kg(9.8+5.0)
tension=740N

3. Oct 8, 2007

### bob1182006

Hm...just curious did you derive f=m(g+x) from F=ma? or was that given to you?

4. Oct 8, 2007

### ideefixem

It wasn't given to me. I assumed it... is it still correct?

5. Oct 8, 2007

### bob1182006

yes, but you should know how to derive it:

since the block is hanging from a rope you know all the forces acting on it, weight and tension of the string.

If you can't see it you can draw a Free Body Diagram, so Tension is pointing upward. weight is pointing down.

using Newton's second law for the y component and using the up direction as positive you will get:

T-mg=ma
T=m(g+a)

because there's only those 2 forces acting on the block, if there were more like multiple strings you'd just use Newton's 2nd law and add up all the forces keeing in mind the +/- direction that you choose.

6. Oct 8, 2007