A uniform chain of length L and mass m is lying on a smooth table and one-nth part of its length is hanging vertically down over the edge of the table. Find the work required to pull the hanging part on to the table.

Let space straight lambda be linear mass density of chain. To pull the chain we have to do work against the weight of hanging part of chain.
Let at any instant length of hanging part be x. 
Therefore the weight of hanging part of chain is 
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#6 {main}</pre>
The work done in pulling the chain by small distance dx is,
                    d W equals negative lambda x g d x

Let  be linear mass density of chain. To pull the chain we have to

Total work done to pull the whole of hanging part of chain is,
             W equals integral d W space equals space integral subscript L divided by n end subscript superscript 0 minus lambda g x space d x
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#6 {main}</pre>
                   space space equals fraction numerator lambda g L squared over denominator 2 n squared end fraction equals fraction numerator m g L over denominator 2 n squared end fraction
                                       

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Show that during a free fall - total energy (Kinetic + Potential) remains constant.


Let a body of mass m be dropped from point A at a height h from the ground.

Let a body of mass m be dropped from point A at a height h from the g
At point A:
        
The kinetic energy is,
                K subscript A italic equals italic 1 over italic 2 m u to the power of italic 2 italic equals italic 1 over italic 2 m left parenthesis 0 right parenthesis squared italic equals 0
and the potential energy is,
                     U subscript A equals m g h
∴     Total energy at point A,
                     E subscript A equals K subscript A plus U subscript A equals 0 plus m g h equals m g h                  ...(1)
At point B:
Let in time t, the body reach point B after falling through a distance x. Let be the velocity of body at B.
 
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#6 {main}</pre>
or                          v squared equals 2 g x
         Kinetic energy at B is,
             K subscript B equals 1 half m v squared equals 1 half m left parenthesis 2 g x right parenthesis equals m g x
 
and potential energy is,
                    U subscript B equals m g left parenthesis h minus x right parenthesis
∴         Total energy at point B is,
            space space space E subscript B equals K subscript B plus U subscript B equals m g x plus m g left parenthesis h minus x right parenthesis equals m g h             ...(2)

At ground:
       
On reaching the ground, let velocity of the body be V.
∴            V squared equals 2 g h
    Kinetic energy at ground is,
             K subscript g equals 1 half m V squared equals 1 half m left parenthesis 2 g h right parenthesis equals m g h
and potential energy is,
              U subscript g equals m g left parenthesis 0 right parenthesis space equals space 0
∴      Total energy at point ground is,
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#6 {main}</pre>
Since total energy at A = total energy at B
                                     = total energy at ground.
therefore total energy during free fall remains conserved.

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Give two illustrations for each of following.
(a) Positive work
(b) Negative work
(c) Zero work.


Positive work:

(i) Work done by gravity on a free falling body is positive.

(ii) Work done by applied force when taken vertically up against gravity is positive.

Negative work:

(i) Work done by friction force on a moving body is negative.

(ii) The work done by gravity on a body moving up is negative.

Zero work:

(i)Work done by electrostatic force of nucleus on electron revolving in a circular orbit is zero.

(ii) Work done by tension force in string on a stone whirled in a circle is zero.

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Derive an expression for potential energy stored in spring.

Consider a massless spring attached with mass m at one end and the other end of spring be connected with a rigid wall. When we pull the mass towards C, the restoring force directed towards A is set up in spring. Against this restoring force we have to do work to displace the mass and hence that work is stored in the form of potential energy in spring.

Consider a massless spring attached with mass m at one end and the ot
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Show that gravitational force is a conservative force.

A force is said to be conservative if work done by the force is independent of the path followed and depends upon the initial and final positions.
Suppose a body of mass m be taken from A to B along different paths as shown in the figure.

A force is said to be conservative if work done by the force is indep

A force is said to be conservative if work done by the force is indep

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