## Homework Statement

Two blocks connected by a string are pulled across a horizontal frictionless surface by a horizontal force of magnitude F=14.0 N. What is the tension (T) in the connecting string?

so you have block 1 (mass of 2kg) then the string then block 2 (mass of 5kg) then the force pulling it of 14 N

---T------->F

f=ma
t=ma

## The Attempt at a Solution

so i found the acceleration:

a = f/m

a = 14/(5+2) = 2 m/s2

then i found the tension:

t = ma

t = 2kg(2m/s2)

t = 4 N

What i am wondering is why do you multiply the 2kg block by the acceleration to get the tension and not multiply the 5kg block or the total of 7kg by the acceleration to get the tension?

can anyone explain please? i have a physics exam tomorrow and i am studying and really trying to understand tension :)
thank you

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

Two blocks connected by a string are pulled across a horizontal frictionless surface by a horizontal force of magnitude F=14.0 N. What is the tension (T) in the connecting string?

so you have block 1 (mass of 2kg) then the string then block 2 (mass of 5kg) then the force pulling it of 14 N

---T------->F

f=ma
t=ma

## The Attempt at a Solution

so i found the acceleration:

a = f/m

a = 14/(5+2) = 2 m/s2

then i found the tension:

t = ma

t = 2kg(2m/s2)

t = 4 N

What i am wondering is why do you multiply the 2kg block by the acceleration to get the tension and not multiply the 5kg block or the total of 7kg by the acceleration to get the tension?

can anyone explain please? i have a physics exam tomorrow and i am studying and really trying to understand tension :)
thank you
You should study free body diagrams (FBD's) first. You solved for the acceleration by looking at the system of both blocks. Good. The only net force acting on the system of 2 blocks is the 14 N force. The tension in the cord between the 2 blocks is internal to the system, and does not enter into the equation when looking at the system of 2 blocks together. Now you then solved for T by isolating the first block in a FBD, using Newton 2. You solved T =4 N, where T is the only force acting in the x direction. Now you could have drawn a FBD of the 2nd block, but then, the NET force acting on the 2nd block alone would be 14-T = ma = (5)(2) =10; then solve T = 4 N. Same result. OK??

yes that makes sense :) thank you

PhanthomJay