What is the tension in a rope for a moving box in Newton's Second Law?

[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.

Homework Statement

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

Homework Equations

a->= F->net/m
(Fnet)x=max
(Fnet)y=may

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

 

[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

 

[bob1182006]

those answers are right.

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

 

[ideefixem]

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

 

[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.

 

[ideefixem]

Thank you for your help.