**Work**– When ever a force makes a body move , then work is said to be done.

- In ordinary language work means almost any physical or mental activity , but in physics it is different
**In physics work is done only when a force produces motion.**- Lets take some example to understand it.
- When an engine pulls a Train the work is done by the engine.
- A horse pulling a cart also doing some work\

So , The work done by a force on a body depends on two factors

**Magnitude of the force****Distance through which the body moves**

So mathematically we can write **Work done = Force X Distance**

Unit of Work As W = F X D = N.m = Joule

Therefore unit of work done is Joule .

Again Let me write the definition for the same

Work done is defined as product of the force and the distance over which the force is applied

The SI unit of work is the joule (J), which is defined as the work expended by a force of one newton through a displacement of one meter.

- Suppose A man wants to push a wall . after trying for sometime he may get exhaust completely. But as there is no change in position involved so the work done is zero.
- Take another example. Suppose you are holding a suitcase in your hand but standing without any movement . In physics the work done is zero. But actually there is involvement of work .

- Suppose A book of mas M kg is lifted upwards to a height of h metre. The work has to be done against gravity.
**so Work done in lifting the body = m X g X h**

W= work done

m= mass of the body

g= acceleration due to gravity

h= height through which the body is lifted

**Problem Time**

- How Much work is done by a force of 20 N in moving an object through a distance of 1 m in the direction of the force.

**We always advise try to solve by your own . at first . then look for the solutions.**

**Solution .**

The work done is calculated s W = F X S ( Force X Distance)

F= 20 N

Distance = s = 1 metere

Work done = 10 * 1 = 10 Joule

**Problem 2**

2. Calculate the work done in lifting 150 kg of water through a vertical height of 10 metere. 9 Assume g = 10 m/s^{2})

As we are lifting the water . so the work is done against gravity .

Hence work = M X g X h ( mass X Acceleration due to gravity X height)

M = 150 Kg

h = 10 Metre

g = 10 m/s^{2}

W= 150 X 10 X 10 = 1500 Joule

**Problem 3**

A Bus has a weight of 2000kg and travelling at a speed of 60m/s stops after travelling 50 meter when the brakes are applied .

What is the force exerted on it by the brakes.

What is the work done by the brakes

**Answer**

In Order to calculate the force ,first we need to find the acceleration(Actually Retardation) first.

Initial Velocity , u = 60 m/s

Final Velocity, v =0 m/s ( As it is stopped at the end)

Distance traveled = 50 metre

recall = v^{2}=u^{2}+2as

0^{2}=60^{2}+2*a*50

100a= -3600

a = -36m/s^{2}

Now Force Applied = m x a = 2000 X 36 = -72000 N

So Force exerted by the brakes on the bus is 72000 newton

work done = Force X Distance

= 72000 X 50 =3600000 Joule = 3.6 X 10^{6} Joule

**Example-4**

A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage

**answer**

Mass of luggage, m = 15 kg

displacement,s = 1.5 m Work done, W = F × s = mg × s

= 15 kg × 10 m s-2× 1.5 m

= 225 kg m s-2 m

= 225 N m = 225 J

Work done is 225 J

- In many cases the movement of the body is at an angle to the direction of applied force
- for example when a child pulls a toy car with a string attached to it the car moves horizontally on the ground but the force applied is at angle to the direction of motion.
**In that case the applied force will be w= Fs Cosθ**- Where θ is the angle between the direction of force and direction of motion.

**Example**

A child pulls a toy car through a distance of 15 metre at an angle of 60^{0 }with the horizontal surface . find the work done if the child applies a force of 10 N.

w= Fs Cosθ

Force = 10 N

Angle θ = 60^{0}

Distnace s = 15 metre

W = 10 X 15 X Cos 60^{0}

= 10 X 15 X 1/2 = 75 Joule

When the force acts perpendicular to the direction of motion or at 90 degree the the work done becomes 0.

w= Fs Cosθ

θ = 90^{0}

Cos 90 = 0

So w= 0

- Work Done is negative when a force acts opposite to the direction of motion of the body
- Example When a football is moving and it stops after some time because of frictional force.
**Here Friction is the negative work as decreases the speed of the ball****Remember the work done when bodies moving around a fixed circular path is zero**.- Example The work done by the earth is 0 during its movement around the sun.

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