Calculate the work done by tension

[ehild]

How does the length L change if the block displaces horizontally by Δx?

ehild

 

[physicslover14]

ldash^2 = 25 +dx^2 + 6dx.
ldash =root(25 +dx^2 + 6dx.)we can ignore dx^2
so ldash =root(25 + 6dx.)
change in length = root(25 + 6dx.)-5

now?

 

[ehild]

What do you think L'-L is if dx is very small?

ehild

 

[physicslover14]

negligible..but there would be some velocity at that point

 

[ehild]

Try. Let dx=0.01, 0.001... Are you familiar with Calculus?ehild

 

[physicslover14]

do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right

 

[ehild]

do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right

Just the opposite dL/dt=3/5 dx/dt. dx/dt = v2, velocity of the 2 kg block, dL/dt = v1, velocity of the 1 kg block.


ehild

 

[physicslover14]

thnx.we could directly say by string constraint no need of calculus.

 

[Tanya Sharma]

ehild...This approach gives the correct result when we consider 2 kg block at point B (which is what is asked in the OP).

But what if , instead of point B ,we had to find speed of 2kg block at any other point(say,point C mid point of the quarter circle AB).Then the length of the string would not have been function of horizontal distance alone.It would have depended on the vertical distance too.

Do we have to use partial differentiation then ?

 

[ehild]

See post #26 and find length L, the hypotenuse of the blue shaded right triangle in terms of the angle theta. Take the derivative with respect to time. V1=dL/dt and v2=Rdθ/dt.

ehild

 

[Tanya Sharma]

See
post #26 and
find length L, the hypotenuse of the blue shaded right triangle in terms
of the angle theta. Take the derivative with respect to time. V1=dL/dt
and v2=Rdθ/dt.

ehild

Thanks...Since my initial response to the problem was incorrect,i have a
doubt whether the net work done on the two blocks is zero ?

 

[ehild]

Thanks...Since my initial response to the problem was incorrect,i have a
doubt whether the net work done on the two blocks is zero ?

You mean the net work of the string, don't you? I think your argument was correct, except the equality of the speeds, and the total work of the tension is zero, but you can derive the work from the forces and displacements.

ehild

 

[Tanya Sharma]

You mean the net work of the string, don't you? I think your argument was correct, except the equality of the speeds, and the total work of the tension is zero, but you can derive the work from the forces and displacements.

ehild

Well.its pretty ironical.I had a correct reasoning about the speeds,but expressed it incorrectly .What I wanted to communicate was that the component of speeds along the direction of length of the string would be equal .

Whereas,I had a wrong understanding of work done by tension ,and wrote the correct result.

Earlier I had a flawed reasoning which prompted me to think that the net work done by the tension would be zero.But now after having a better perspective ,i am not convinced that how the work done by tension on 1 kg block would be equal and opposite to the work done by tension on 2 kg block.Afterall they cover unequal distances,with varying angle between tension and displacements in case of 2 kg block.

Why would the net work done by tension be zero ?

I had read somewhere that the net work done by the internal forces is zero .Is this a general statement or applicable in special circumstances ?

 

[haruspex]

i am not convinced that how the work done by tension on 1 kg block would be equal and opposite to the work done by tension on 2 kg block.

The tensions each side of the pulley are the same at all times, the distance moved by the string in each elemental time interval is in the direction of the corresponding tension (except one is negative), and the total distance moved by the string is the same each side. So the integral must be equal and opposite.

 

[haruspex]

thnx.we could directly say by string constraint no need of calculus.

How did you get the 3/5 without using calculus - or something equivalent?

 

[Tanya Sharma]

How did you get the 3/5 without using calculus - or something equivalent?

The component of velocity of 2 kg block along the direction of length of string will be equal to the component of velocity of 1 kg block along the length of string.

Let θ be the angle between the velocity of 2 kg block (V₂) and the direction of length of string.

When 2Kg block is at B ,V₂ cosθ = V₁ ,where cosθ =3/5.

Hence,V₁/V₂ =3/5 .

This works in this case nicely as we had the angle θ known .

 

[Tanya Sharma]

The tensions each side of the pulley are the same at all times, the distance moved by the string in each elemental time interval is in the direction of the corresponding tension (except one is negative), and the total distance moved by the string is the same each side. So the integral must be equal and opposite.

The tension acting at the 2 kg block is not tangential at every instant,whereas the displacement is tangential to the arc.The angle between tension and the displacement changes continuously.The displacement is surely not in the direction of tension in case of 2 kg block

Yes, the tension is same on both the blocks.But I still can't figure out why the work done on one block will be equal and opposite to the other block.

 

[ehild]

Why would the net work done by tension be zero ?

I had read somewhere that the net work done by the internal forces is zero .Is this a general statement or applicable in special circumstances ?

The block moves along an arc of circle, its displacement is not parallel with the string. The tension acts at the end of the string, and the end moves along the arc. At small displacement Δr the work is the dot product δW=TΔr, T multiplied by the component of the displacement along the string. The component of displacement along the string is the same for both blocks, as the length of the string does not change, but the forces act in opposite directions, so the net work is zero.

The work of internal forces is not zero in general. Think of two masses connected by a spring. Let they be equal. The spring is stretched, then relaxed. Both masses move inward by Δx. The spring does positive work on both masses, the net work is 2(1/2 kΔx²).

ehild

 

[Tanya Sharma]

The block moves along an arc of circle, its displacement is not parallel with the string. The tension acts at the end of the string, and the end moves along the arc. At small displacement Δr the work is the dot product δW=TΔr, T multiplied by the component of the displacement along the string. The component of displacement along the string is the same for both blocks, as the length of the string does not change, but the forces act in opposite directions, so the net work is zero.


ehild

Yes...that makes complete sense . Brilliant!