Tension on a rope between two objects with different mass

[siumbaim]

Homework Statement

A 3kg block is connected to a 5kg block by a rope. A second rope is attached to the 5kg block and is being pulled by someone with a force of 11N. What is the tension on the rope between the two blocks? No friction

Homework Equations

F=ma[/B]

The Attempt at a Solution

So I calculated the acceleration for the 5kg block with the 11N and it was 2.2m/s^2. I assumed that since the 5kg block is accelerating at 2.2m/s^2, the 3kg block must be accelerating at the same value. So then I did, F=ma,
F=(3kg)(2.2m/s^2)= 6.6N. So the tension on the rope between the two objects is 6.6N. Did I do this right? I don't have the solutions for this. Thank you!

 

[Dick]

No, not right. You are right that the acceleration of both blocks is the same. But then the acceleration would be that which is produced by an 11N force acting on the combination of the two blocks.

 

[siumbaim]

No, not right. You are right that the acceleration of both blocks is the same. But then the acceleration would be that which is produced by an 11N force acting on the combination of the two blocks.

Ok so you know how if you drop two objects with different masses, they would fall at the same time in a system with no friction or anything... because gravity is constant. Is this question kind of like that in a way? Would the tension between the two objects be 11N?

 

[Dick]

Ok so you know how if you drop two objects with different masses, they would fall at the same time in a system with no friction or anything... because gravity is constant. Is this question kind of like that in a way? Would the tension between the two objects be 11N?

No, the 11N force is acting on a system with a total mass of the sum of both masses. What's the common acceleration of the two masses?

 

[siumbaim]

No, the 11N force is acting on a system with a total mass of the sum of both masses. What's the common acceleration of the two masses?

common acceleration is a=(11N)/(8kg)=1.375m/s^2. Soooooo F=(1.375m/s^2)(3kg) =4.125N? Is that the answer?

 

[Dick]

common acceleration is a=(11N)/(8kg)=1.375m/s^2. Soooooo F=(1.375m/s^2)(3kg) =4.125N? Is that the answer?

Right. The only force acting on the 3kg block is the tension in the rope. So that must be the tension.

 

[siumbaim]

Right. The only force acting on the 3kg block is the tension in the rope. So that must be the tension.

THANKS!

 

[Hungry Hippo]

Used this as a reference for my AP Physics homework, helped a lot. Thanks guys

 

[haruspex]

4.125N?

Yes, except that you could lose marks for quoting too many significant figures.

 

[Chestermiller]

It might have helped a little on this problem if you had drawn some free body diagrams. You know what free body diagrams are, correct?