#### Physics (NTSE/Olympiad)

# 3. Work and Energy

__Work Done Analysis__

**Work done when force and displacement are along same line.**

**Work done by a force :** Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.

**Work done against the force : ** Work is said to be done against a force if the direction of the displacement is opposite to that of the force.

**Work done against Gravity : **To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If 'm' is the mass of the object and 'h' is the height through which it is raised, then the upward force

(F) = force of gravity = mg

If 'W' stands for work done, then

W = F . h = mg . h

Thus W = mgh

Therefore we can say that, "The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted.

** Example : 3**

Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.

** Solution. **Force, F = 120 N; Distance, s = 100 m

Using the formula, we have

W = Fs = 120 N × 100 m = 12,000 J

** Example : 4**

A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s^{2}. Calculate the work done.

** Solution. **Given :

mass, m = 5 kg

acceleration, a = 3 m/s^{2}

Force acting on the body is given by

F = ma = 5 × 3 = 15 N

Now, work done is given by

W= Fs = 15 N × 4 m = 60 J

** Example : 5**

Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms^{–2}).

** Solution. **Force of gravity

mg = 200 × 9.8 = 1960.0 N

h = 5 m

Work done, W = mgh

or W= 1960 × 5 = 9800 J