Question

A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.1 kg. A force of magnitude F = 26.0 N pulls m1 to the left a distance d = 0.83 m.

How much work is done by the tension (in-between the blocks) on block m2?

What is the tension in the string?

What is the NET work done on m1?

Answer 1

The second tension in the string question I'm assuming is for the string pulling the blocks, not the one in the middle.

Answer 2

I dont think it is the string on the corner. If it is, there is nothing calculate it is simply 26N. I think they are asking for the tension of the rope in the middle.

Find the tension of the middle string by applying F=ma to the system, then to either the mass m1 or m2

Answer 3

W = 26*.83 .. which is 21.58 N...

Answer 4

Force is not 26N in the rope in the middle. What you've found is the work done by the tension in the rope on the corner side

Answer 5

I see, the answer isnt 26... so idk.

Answer 6

I've given the steps above why don't you try out that?

Answer 7

I did that originally and neither worked.

Answer 8

Which is why I posted the question on this site :)

Answer 9

Okay I'll help a bit further,
take the tension of the end rope as F_1 and the the tension of the middle rope T and the acceleration of the system is -a

As I've mentioned above
by applying F=ma to the system,
$$F_1=(m_1+m_2)a\\
a=\frac{F_1}{m_1+m_2}$$

apply F=ma to m_2,
$$T=m_2a\\
T=\frac{m_2F_1}{m_1+m_2}$$

Can you find the Tension of the middle rope now?

Answer 10

yes, I did that. a = 2.11
and when plugged into the T = m2a it is 4.7*2.11 which is 8.6 which is not the right answer. 4.7*21.58/9.6 = 4.74 which also is not the answer.

Thank you for your help but this was what I did originally with no luck.