Intuitive explanation of why work done by tension is 0?

[navneet9431]

Homework Statement

Homework Equations

Work Done=Force*Displacement in the direction of the force

The Attempt at a Solution

I tried to solve the problem this way,
I wrote these equations for the two masses,
T-mg=ma...(i)
Mg-T=Ma...(ii)
From (i)&(ii), T=[M(g-a)+m(a+g)]/2
and then I would multiply them with the displacement to get the net work done 0.
So this is how I solved it mathematically.
Can you please provide an intuitive explanation of why the net work done by tension would be zero?
I will be thankful for help!

 

[phinds]

Can you please provide an intuitive explanation of why the net work done by tension would be zero?

Suppose you have two tug-of-war teams pulling on a rope and both teams are at a standstill. What work is being done?

 

[navneet9431]

But the pulley is not standstill.right?

Suppose you have two tug-of-war teams pulling on a rope and both teams are at a standstill. What work is being done?

 

[phinds]

But the pulley is not standstill.right?

OK, suppose one team walks forward at 2mph and the other team walks backwards at 2mph, it which case the tension is unchanged. How much work does the tension do?

 

[haruspex]

provide an intuitive explanation of why the net work done by tension would be zero?

For a constant force, work = force times displacement.
In a tug of war, equal and opposite forces, equal displacements. Fd+(-F)d=0.
For the pulley set up in the diagram, equal forces, equal and opposite displacements. F.d+F(-d)=0.
Easy to generalise to variable forces using integrals.

 

[navneet9431]

For a constant force, work = force times displacement.
In a tug of war, equal and opposite forces, equal displacements. Fd+(-F)d=0.
For the pulley set up in the diagram, equal forces, equal and opposite displacements. F.d+F(-d)=0.
Easy to generalise to variable forces using integrals.

Thanks!
It was easy to understand.