Here, m = 40 kg, g = 10 ms^-2, h = 5m
Potential energy of the object at a height of 5m,
E_p = mgh =40 x 10 x 5= 2000J
When the object is half-way down, let its velocity be v.Then
v²- u² = 2gs
v² - 0² = 2 x 10 x 2.5 or v² = 50
Kinetic energy of the object, E_k=1/2 x 40 x 50 = 1000 J

m = 20 kg, u = 5 ms^-1, v = 2 ms^-1
Work done=Change in K.E.
W = 1/2 mv²-1/2mu² = 1/2m(v²-u²)
= 1/2 x 20(2²-5²) = 10 (4-25) = - 10 x 21
= - 210 J
The negative sign indicates the retarding nature of the applied force.

(i) Yes, Suma is doing work by pushing water in the backward direction.
(ii) No, because the force exerted by donkey in the upward direction is perpendicular to the horizontal displacement of the load.
(iii) Yes, work is done in lifting water against the force of gravity.
(iv) No, because the leaves of plants remain at rest during photosynthesis.
(v) Yes, engine is doing work in pulling the train.Both the applied force and displacement are in same direction.
(vi) No, because food grains reamins at rest.
(vii) Yes, work is done by the wind in moving the sailboat.

(i) Energy consumed by a 100 W bulb each day
= 100 W x 2 h=200 Wh = 200/1000 = 0.2 kWh
Energy consumed by five 40 W tubes each day
= 5 x 40 W x 4 h = 800 Wh = 800/1000 = 0.8 kWh.
Total energy consumed each day
= 5 x 40 W x 4 h = 800 Wh = 800/1000 = 0.8 kWh
Total energy consumed each day = 0.2 + 0.8 = 1.0 kWh
Total energy consumed in 60 days = 1.0 x 60 = 60 kWh
(ii) Cost of 1 kWh =  Rs. 3
Cost of 60 kWh = 3 x 60 =  Rs.180

The SI unit of power is watt(W). The power of an agent is one watt if it does work at the rate of 1 joule per second.
1 watt = 1joule/1 second. or 1W = 1 Js^-1
We can also say that power is 1 watt when the rate of consumption of energy is 1 Js^-1 .
Bigger units of power are kilowatt, megawatt, and horsepower.
1 kilowatt = 1000 watt or 1 kW = 1000 W
1 megawat t = 10⁶ watt or 1 MW = 10⁶ W
1 horse power =746 watt or 1 H.P = 746 W

Here m = 25 kg, h = 10m, g = 10m/s²
(a) Work done = mgh =25 x 10 x 10J = 2500J
(b) Gravitational P.E.= Work done  =2500J
(c) Gravitational P.E. at a height of 30m = 25 x 30 x 10J = 7500J
P.E. at 30 m height/P.E. at 10m height = 7500J/2500J = 3
Thus, P.E. increases by 3 times.
(d) K.E. of the body will be maximum just before the body strikes the ground.

Mass of the car, m = 1500 kg
Initial velocity of the car, 
u = 30 km h^-1= 30 x 1000 m/3600 s = 25/3 ms^-1
Final velocity of the car, v = 60 km h^-1= 60 x 1000m / 3600 s = 50/3 ms^-1
Work done on the car = Increase in K.E. of the car = Final K.E. - Initial K.E.
= 1/2m (v²-u²) = 1/2 x 1500 [(50/3)² - (25/3)²]J
= 1/2 x 1500 (50/3 + 25/3) (50/3 - 25/3)J = 1500/2 x 75/3 x 25/3J = 156250J

Energy of a body is defined as its capacity or ability to do work.When a body is capable of doing more work, it is said to possess more enrgy.
Examples:
(i) When a fast moving cricket ball hits a stationary wicket, it throws away the wicket.The moving ball possess energy and hence does work on the wicket.
(ii) A hammer raised through a certain height possesses energy.When it falls on a nail placed on a piece of wood, it drives the nail into the wood.
(iii) When a child winds a toy car and does work in moving against the force of friction.
In all the above examples, the objects acquire capability of doing work through different means and these objects are said to possess energy.

Here m = 1000 kg, u=30 m/s, v = 0, s = 50 m
v² - u² = 2as
0-30² = 2a.50 or a = -900/100 = -9 m/s²
Deceleration=  9m/s²
decelerating force, F = ma =1000 x 9 = 9000 N
Work done by the brakes, W = F.s = 9000 x 50J = 450000J