Question

Solve the following examples.

a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?

b. If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?

c. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?

d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg?

e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi?

Answer 1

a. Power, P = 2 kW
$$\begin{gathered} \mathrm{Power},P=\frac{\mathrm{Work}\mathrm{done}\left(W\right)}{\mathrm{Time}\mathrm{taken}\left(T\right)} \\ \Rightarrow W=P\times T \end{gathered}$$
For every minute i.e. 60 s, work done by the pump is
$W=2\times 1000\times 60=120000\mathrm{J}$
Now, this work done is stored in water as its potential energy. Thus,
$$\begin{gathered} mgh=120000\mathrm{J} \\ \Rightarrow m=\frac{120000}{9.8\times 10}=1224.5\mathrm{kg} \end{gathered}$$
Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.

b. Total electricity consumed in the month of April is
$$\begin{gathered} E=P\times T\times (\mathrm{Total}\mathrm{days}\mathrm{in}\mathrm{April}\mathrm{month}) \\ E=1200\times 30\times 60\times 30=64.8\times {10}^6\mathrm{J} \\ \mathrm{We}\mathrm{know}, \\ 1\mathrm{unit}=3.6\times {10}^6\mathrm{J} \\ \mathrm{Thus}, \\ E=18\mathrm{units} \end{gathered}$$

c. Energy of the ball at the height of 10 m = mgh = 10mg
Let the ball rebounds to a height h' where the energy reduces by 40%. Thus,
Energy at height h' = mgh' = 60% of energy at height of 10 m = $\frac{60}{100}\times 10mg=6mg$
or, mgh' = 6mg = 6 m

d. Here, v = 72 km/h = 20 m/s, u = 54 km/h = 15 m/s, m = 1500 kg
Work done by the car = Change in kinetic energy of the car
i.e.
$$\begin{gathered} W=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{i}} \\ W=\frac{1}{2}\times m\times v^2-\frac{1}{2}\times m\times u^2=\frac{1}{2}\times m(v^2-u^2) \\ =\frac{1}{2}\times 1500({20}^2-{15}^2)=131250\mathrm{J} \end{gathered}$$

e. Work done, W = F$\times$S$\times$cos $\theta$
Here, $\theta$ = 0^o
F = 10 N
S = 30 cm = 0.03 m
$\therefore W=10\times .03\times 1=3\mathrm{J}$