**Problem Statement:**

We have a chain lying on the ground, with uniform linear density λ. It is held by one of its end and is pulled vertically upwards with a constant velocity, v

1.)The Force in terms of the length (l) of the end pulled.

2.)Change in energy up to that point.

3.)Power

We have a chain lying on the ground, with uniform linear density λ. It is held by one of its end and is pulled vertically upwards with a constant velocity, v

_{0}. So calculate the following :-1.)The Force in terms of the length (l) of the end pulled.

2.)Change in energy up to that point.

3.)Power

Solution. 1) By differentiating momentum with respect to time, we get the force applied.

[tex]F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}[/tex]

[tex]F = \lambda lg + \lambda v_0^2[/tex]

Power, as calculated by the Force,

[tex]P = Fv_0[/tex]

[tex]P = \lambda lgv_0 + \lambda v_0^3[/tex]

2.) Kinetic Energy – Every infinitesimally small element of the chain off the ground is moving with same velocity v_{0}. ∴

[tex] K = \frac{1}{2}\lambda lv_0^2[/tex]

Potential Energy, calculating the PE for infinitesimally small element of length dx at height x.

[tex] dU = \lambda dx.g.x[/tex]

[tex] U = \int_{0}^{l}\lambda dx.g.x = \frac{1}{2} \lambda gl^2[/tex]

Power from Energy,

[tex]P = \frac{dE}{dt} = \frac{dK}{dt} + \frac{dU}{dt}[/tex]

[tex]P = \frac{1}{2}\lambda v_0^2\frac{dl}{dt} + \frac{1}{2}\lambda g\frac{dl^2}{dt}[/tex]

[tex]P = \frac{1}{2}\lambda v_0^3 + \lambda glv_0[/tex]

My question is, the power calculated from the energy is less than the power from Force. Any ideas what that imply?

http://ift.tt/1jBy4yf