- #36
physicslover14
- 16
- 0
do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right
physicslover14 said:do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right
Thanks...Since my initial response to the problem was incorrect,i have aehild said: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 said: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 said: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
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.Tanya Sharma said: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.
How did you get the 3/5 without using calculus - or something equivalent?physicslover14 said:thnx.we could directly say by string constraint no need of calculus.
haruspex said:How did you get the 3/5 without using calculus - or something equivalent?
haruspex said: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.
Tanya Sharma said: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 ?
ehild said: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