# Work, energy and power

Work:

Work is defined as the product of the magnitude of the displacement and the component of the force in the direction of the displacement. W=F.d = Fdcos$\theta$

Work is scalar quantity because it is defined as the dot product or scalar product two vectors force and displacement and the scalar product of two vectors gives the scalar quantity.

W = $\vec F. \vec D$

Potential energy:

The energy possessed by a body due to its position in a force field is called potential energy.  P.E=mgh

Kinetic energy and the momentum of an object:

Kinetic energy is possessed by a body due to its motion whereas momentum is product of mass and velocity. When the shot is fired from a canon the change in momentum is zero because momentum of the system remains conserved but the kinetic energy increases because the chemical energy is converted into the mechanical energy.

Relation between watt and horsepower:

Both horse power and kilowatts are units of power.

Power (kW) =power (HP) * 0.7

Power (HP) = Power (kW) * 1.341

Conservative and non – conservative forces:

A force is said to be conservative if the work done by or against the force in moving body depends on the initial and final positions of the body and independent on the nature of the path followed.

The earth-moon system is conservative. The work from point a to b then point b to a is zero. The moon follows an elliptical orbit. The energy at some specific point in said orbit remains the same. As the distance changes so does the energy. But when the moon returns to the same point it was at previously, it has exactly the same energy as previously so there is no change of energy taking place.

A force is said to be non-conservative if the work done by or against the force in moving body depends on nature of the path followed between these points.

Frictional forces are non-conservative forces .let us consider a body is moved from position A to B .work done in moving this body from A to B does not only depend upon the position A and B but also on the length

An expression to calculate the work done by a variable force;

Work is said to be done on the body if the external force applied to the body displace it in any direction except perpendicular to the direction of force.

Let variable force acting on a body to displace it from A to B in a fixed direction. A to B is consists large number of infinitesimal displacement. Let’s take PQ = dx infinisitmely small displacement and along this displacement force is constant in magnitude as well as in direction.

Small amount work done in moving the body from P to Q is

Dw = F * dx  = PQ *PS  =  area of PQRS

Total work done from A tto B = W = Σ dW = Σ Fdx

It can be written as

W =$\mathop {\lim }\limits_{{\rm{dx}} \to 0{\rm{\: }}} \Sigma {\rm{\: Fdx}}$  = $\mathop \smallint \limits_{{{\rm{x}}_{\rm{A}}}}^{{{\rm{X}}_{\rm{B}}}} {\rm{Fdx}}$ =  $\mathop \smallint \limits_{{{\rm{x}}_{\rm{A}}}}^{{{\rm{X}}_{\rm{B}}}} {\rm{area\: of\: PQRS}}$  = Area of ABCDA

∴ Work done by the variable force is equal to the area under the force curve and the displacement axis.

Potential energy:

The energy possessed by the body due to its position or configuration is called potential energy.eg. a stretched rubber band. Let us consider a body of mass m on the surface of the earth. The earth attracts it with the force equal to its weight mg. If it is taken to height h then the amount of work done against the gravity is W = mg *h.  Hence the work done is stored in the form of gravitational potential energy.

Principle of conservation of the energy:

Principle of conservation of the energy states that,” the for an isolated system the total energy of the system remains constant”.

Let us consider a body of mass m falling freely under the effect ogf gravity from the height h as shown in figure.

At point A (initially) body is at rest

K.E of body = 0

P.E of the body = mgh

Total energy of the body = K.E + P.E = mgh

At point B the body is at height of (h-x ) from the ground .

P.E of the body = mg (h-x)

If v is the velocity at point B then

V2  = U2 + 2as

V2  = 2gx

∴ K.E of the body =  $\frac{1}{2}{\rm{m}}{{\rm{v}}^2} = {\rm{mgx\: }}$

Total energy of the body = K.E + P.E = mgh

At point C

Id v is the velocity just before striking the ground

${{\rm{V}}^2} = 0 + 2{\rm{gh}}$

${{\rm{v}}^2} = 2{\rm{gh}}$

K.E = $\frac{1}{2}{\rm{m}}.2{\rm{gh}}$  = mgh

P.E = mgh = 0

Total energy of the body = K.E + P.E = mgh

This shows that sum of P.e and K.E of the falling bodied remain constant

An elastic collision in one dimension, the relative velocity of approach before the collision is equal to relative velocity of separation after collision:

Consider two particles, of m1 and m2, the masses, u1 and u2 the velocities before collision, and v1 and v2 the velocities after collision.

The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision, and is expressed by the equation

${{\rm{m}}_1}{{\rm{u}}_1} + {\rm{\: }}{{\rm{m}}_2}{{\rm{u}}_2}{\rm{\: }} = {{\rm{m}}_1}{{\rm{v}}_1} + {\rm{\: }}{{\rm{m}}_2}{{\rm{v}}_2}$

${{\rm{m}}_1}{\rm{\: }}\left( {{{\rm{u}}_1} - {{\rm{v}}_1}} \right) = {\rm{\: }}{{\rm{m}}_2}{\rm{\: }}\left( {{{\rm{v}}_2} - {{\rm{u}}_1}} \right){\rm{\: }}$………..1

${\rm{\: }}$

Also the kinetic energy before and after collision are equal

$\frac{1}{2}{{\rm{m}}_1}{\rm{u}}_1^2{\rm{\: }} + {\rm{\: }}\frac{1}{2}{{\rm{m}}_2}{\rm{u}}_2^2{\rm{\: }} = {\rm{\: }}\frac{1}{2}{{\rm{m}}_1}{\rm{v}}_1^2 + \frac{1}{2}{{\rm{m}}_2}{\rm{v}}_2^2$

${{\rm{m}}_1}{\rm{\: }}\left( {{{\rm{u}}^2}_1 - {\rm{v}}_1^2} \right) = {\rm{\: }}{{\rm{m}}_2}{\rm{\: }}\left( {{\rm{v}}_2^2 - {\rm{u}}_1^2} \right)$………..2

Solving 1 and 2 we get

${{\rm{u}}_1} + {\rm{\: }}{{\rm{v}}_1} = {{\rm{v}}_2} + {\rm{\: }}{{\rm{u}}_2}$

${\rm{\: }}{{\rm{u}}_1} - {{\rm{u}}_2} = {{\rm{v}}_2} - {{\rm{v}}_1}$

This shows that the relative velocity of approach before the collision is equal to relative velocity of separation after collision.

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