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Work, Energy and Power

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Physics Part I

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Physics

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Class 10 Class 12
Is work a scalar or a vector quantity?

Work is dot product of two vectors. i.e. F with rightwards arrow on top times S with rightwards arrow on top.

And dot product is a scalar quantity. Therefore, work is a scalar quantity. 
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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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The work done in pulling the chain by small distance dx is,

                    d W equals negative lambda x g d x

Total work done to pull the whole of hanging part of chain is,

             W equals integral d W space
space space space equals space integral subscript L divided by n end subscript superscript 0 minus lambda g x space d x 

               equals right enclose negative lambda g x squared over 2 end enclose subscript bevelled L over n end subscript superscript 0

              space space equals fraction numerator lambda g L squared over denominator 2 n squared end fraction
space equals fraction numerator m g L over denominator 2 n squared end fraction
                                       

1531 Views

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. Work has to be done against this restoring force in order to displace the mass and hence this work is stored in the form of P.E in the spring. 

Let at any instant, mass m be at a distance x from A. 

Restoring force at this instance is, 

                      straight F with rightwards harpoon with barb upwards on top subscript r space equals space minus space k x space
Therefore, to keep the mas in equilibrium, we have to apply the force Fa equal and opposite to -Fr.

If the mass is further displaced by dx with rightwards harpoon with barb upwards on top, the amount of work done dW for this dispalcement by applied force is,

dW space equals space straight F with rightwards harpoon with barb upwards on top subscript straight a space dx with rightwards harpoon with barb upwards on top space

space space space space space space equals space straight F subscript straight a space dx space

space space space space space space equals space straight k space straight x space dx


Total amount of work done in order to displace the mass from mean position of A to C is using applied force Fa is, 

straight W space equals space integral subscript straight A superscript straight C dW space equals space integral subscript 0 superscript straight a straight k space straight x space dx space equals space 1 half kx squared vertical line subscript 0 superscript straight a space equals space 1 half ka squared space

Work done by applied force against this restoring force is stored in the form of potential energy is, 

                straight W space equals space straight P space equals space 1 half ka squared
1082 Views

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.

 
i) Work done if body is taken up straight along Ab, 

                 W1 = -mgh

ii) Work done by gravity if body is taken up along path ACB, 

 W2 = WAC + WCB 

straight W subscript AC space equals space straight m space straight g with rightwards harpoon with barb upwards on top. space AC with rightwards harpoon with barb upwards on top space

space space space space space space space space space space equals space minus space mg space left parenthesis AC right parenthesis space cos space left parenthesis space 90 space minus space straight theta right parenthesis

space space space space space space space space space space space equals space minus mg space AC space sin space straight theta space
space space space space space space space space space space space equals space space minus space mg space left parenthesis AB right parenthesis space equals space minus space mgh space

straight W subscript BC space equals space straight m space straight g with rightwards harpoon with barb upwards on top. space CB with rightwards harpoon with barb upwards on top

space space space space space space space space space equals space minus mg space left parenthesis CB right parenthesis space cos space left parenthesis 90 right parenthesis space equals space 0

Therefore comma space

straight W subscript 2 space equals space minus space mgh space
 
iii) Work done by the gravity along the path ADEFGHB, 

W3 = WAD + WDE + WEF + WFG + WGH + WHB 

 Here comma space the space displacements space DE with rightwards harpoon with barb upwards on top comma space FG with rightwards harpoon with barb upwards on top space and space JB with rightwards harpoon with barb upwards on top space are space
perpendicular space to space gravity. space

Therefore comma space

straight W subscript DE space equals space straight W subscript FG space equals space straight W subscript HB space equals space 0 space comma space and space

straight W subscript AD space equals left parenthesis space straight m. straight g with bar on top right parenthesis space. space AD with rightwards harpoon with barb upwards on top space equals space minus space mg space left parenthesis AD right parenthesis space

straight W subscript EF space equals space left parenthesis space straight m. straight g with bar on top right parenthesis space. space EF with rightwards harpoon with barb upwards on top space equals space minus space mg space left parenthesis EF right parenthesis space

straight W subscript GH space equals space left parenthesis space straight m. straight g with bar on top right parenthesis space. stack GH space with rightwards harpoon with barb upwards on top equals space minus space mg space left parenthesis GH right parenthesis space

Therefore comma space

straight W subscript 3 space equals space minus space mg space left parenthesis AD space plus space EF space plus space GH right parenthesis space equals space minus space mgh space

Since space straight W subscript 1 space equals space straight W subscript 2 space equals space straight W subscript 3 space

Therefore comma space gravitational space force space is space conservative. space


3289 Views

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.

 

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

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.
 

Now,

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rightwards double arrow                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
 
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 

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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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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