Work On Two Blocks

For this, you will need to know that:

\[ work=force \times displacement \]

force: f
displacement: s
work: w
tension: t

Except there is a force acting on m1 because it is connected to m2 via a string, so we will subtract the tension from 37 N (given in the question). We subtract because the tension is going in the opposite way relative to the left where is moved:

\[ work = (f-t)\times d \]
\[ work = (37-t)\times 0.7 \;\;\;\;\; (1) \]

so now we need to find the tension...

Remember that tension is the same as force, so it can be calculated used newton's second law

\[F=ma\]

\[T=ma\]

\[a = \frac{T}{m} \]

\[ \frac{37}{5.4} = \frac{T}{5.4} - \frac{T}{4.5} \]

Solve for \(T\), you get:

\[T= 16.8 N \]

Now go back to equation \((1)\) and sub this in:

\[ work = (37-16.8)\times 0.7 = 14.1 N \]

Therefore, the net work done on \(m_1\) is 14.1 N

Forgive me for how this will likely sound, but would the last one not be in joules rather than Newtons?

Post Merge: 2 years ago

Also, that helped a ton, thank you immensely for your aid. I still am not sure how to figure out how much work tension does on mass 2. I think I do not understand how to account for it´s mass if that makes sense.

I apologize for how stupid I likely sound. I have been learning things before the teacher teaches us due to personal circumstances.

Yes, my bad... \(N \cdot m\) is a Joule. Yikes

I'm not entirely sure.

I believe the tension due to both strings is equal to the magnitude of the force, 37N.

Could you try 18.5 N times the 0.73 m?

Sorry to mislead you if you get it wrong.

That is not correct, no. But thank you for the rest of your help!

Dang, as you can tell, physics isn't my forte. Feel free to post any further questions

For m1 :

=> T = m1 a

=> T = 6a ------- (1)

For m2 :

m2g - T = m2a

=> 3.3 (9.8) - T = 3.3 a -------- (2)

add up (1) and (2) u get :

=> 3.3(9.8) = 9.3 a

=> a = 3.48 m/s^2

from (1) T = 6a = 20.88 N

so W = Td = 20.88 (0.87) = 18.17 J <<<<<<<<