Class 9 Work and Energy Notes

Notes Important Questions

Work and Energy Notes

Work:-

Work is said to be done if a force is applied on an object & it displaces in the direction of force applied.

  • S.I. unit:- Joule (J)
  • It is a scalar quantity.
  • Work done = Force x displacement (W = FS)

1 Joule :- The work done on an object is said to be 1 joule when a force of 1 N displaces it by 1 m along the line of action of the force.

Positive, Negative and Zero Work Done:-

  • W = F S Cos θ

Energy:-

The capacity of doing work is called energy.

  • S.I. unit :- Joule (J)

Kinetic energy:-

The energy possessed by an object by virtue of its motion is known as kinetic energy. Ex:- Energy in moving fan.

Expression of kinetic energy:-

Consider an object having mass ‘m’ moving with uniform acceleration ‘a’. Let its initial velocity be ‘u’ & final velocity be ‘v’.

From 2nd law of motion, F = ma

W = FS

or, W = ma(v2-u2)/2a [Since, v2-u2 = 2aS So, S = (v2-u2)/2a]

or, W = m(v2-u2)/2 = ½ m(v2-u2)

If u = 0 then, W = ½ m(v2-02) = ½ mv2

So, Ek = ½ mv2

Work done= Change in kinetic energy

or, W = Ekf -Eki

or, W = ½ mv2½ mu2

Potential energy:-

The energy possessed by an object by virtue of its position or configuration is known as its potential energy. Ex:- Water stored in a dam.

Expression of potential energy:-

Consider an object having mass ‘m’ at a height ‘h’.

From 2nd law of motion,

F = ma = mg (Where, g is acceleration due to gravity.)

Now, W = FS

or, W = mgh [here, S = h]

So, Ep= mgh

Law of conservation of energy:-

It states that energy can neither be created nor be destroyed. It can only be converted from one form to another.

Mechanical energy:-

The sum of kinetic energy & potential energy of an object is known as mechanical energy.

  • Mechanical energy = Potential energy +Kinetic energy = constant

Mathematical proof of conservation of mechanical energy for a freely falling object:-

Consider an object having mass ‘m’ at height ‘h’.

At point A, velocity = 0, acceleration = g and height = h

K.E. = ½ mv2 = ½ m.02 = 0

P.E. = mgh

M.E. = K.E.+P.E. = 0 + mgh = mgh

At point B,

Using, v2-u2 = 2aS

or, v2 = 02+2gx [here,u=0, a=g & S=x]

or, v2 = 2gx

K.E.= ½ mv2 = ½ m.2gx = mgx

P.E.= mg(h-x) = mgh – mgx

So, M.E.=K.E.+P.E. = mgx + mgh – mgx = mgh

At point C,

Using, v2-u2 = 2aS

or, v2= u2+2aS = 02+2gh = 2gh

K.E.= ½ mv2 = ½ m(2gh) = mgh

and P.E.= mgh = mg.0 = 0

so, M.E.= K.E.+P.E. = mgh+0 = mgh

Clearly, the total mechanical energy for a freely falling object is ‘mgh’ i.e., constant at every point.

Power:-

The rate of doing work is called power.

  • Power= work/time (P=W/t)
  • S.I. unit :- J/S or Watt (W)

1 watt of power:- The power of an agent is said to be 1 watt if it does 1 J work in 1 sec.

  • 1 kW = 1000 W = 1000 J/s
  • Commercial unit of energy :- kilowatt hour (kW h)

Conversion of commercial unit of energy to S.I. unit of energy:-

1 kwh = 1000 W x 3600 sec = 3.6 x106 W sec = 3.6 x106 J

or, 1 unit = 1 kwh = 3.6 x 106 J

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