Class 9 RATE OF DOING WORK

### Topic Covered

color{red} ♦ RATE OF DOING WORK

### RATE OF DOING WORK

Do all of us work at the same rate? Do machines consume or transfer energy at the same rate? Agents that transfer energy do work at different rates. Let us understand this from the following activity:

Activity ____________11.16

♦ Consider two children, say A and B. Let us say they weigh the same. Both start climbing up a rope separately. Both reach a height of 8 m. Let us say A takes 15 s while B takes 20 s to accomplish the task.

♦ What is the work done by each?

♦ The work done is the same. However, A has taken less time than B to do the work.

♦ Who has done more work in a given time, say in 1 s?

A stronger person may do certain work in relatively less time. A more powerful vehicle would complete a journey in a shorter time than a less powerful one.

We talk of the power of machines like motorbikes and motorcars. The speed with which these vehicles change energy or do work is a basis for their classification.

Power measures the speed of work done, that is, how fast or slow work is done. Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by:

Power = work/time

or  P = W/t ..........(11.8)

The unit of power is watt having the symbol W. 1 watt is the power of an agent, which does work at the rate of 1 joule per second. We can also say that power is 1 W when the rate of consumption of energy is 1 J s^(–1).

1 watt = 1 joule/second or 1 W = 1 J s^(–1). We express larger rates of energy transfer in kilowatts (kW).

1 kilowatt = 1000 watts
1 kW = 1000 W
1  kW = 1000 J s^(–1).

The power of an agent may vary with time. This means that the agent may be doing work at different rates at different intervals of time. Therefore the concept of average power is useful. We obtain average power by dividing the total energy consumed by the total time taken.

### COMMERCIAL UNIT OF ENERGY

The unit joule is too small and hence is inconvenient to express large quantities of energy. We use a bigger unit of energy called kilowatt hour (kW h).

What is 1 kW h? Let us say we have a machine that uses 1000 J of energy every second. If this machine is used continuously for one hour, it will consume 1 kW h of energy. Thus, 1 kW h is the energy used in one hour at the rate of 1000 J s^(–1) (or 1 kW).

1 kW h = 1 kW xx 1 h

= 1000 W xx 3600 s

= 3600000 J
1 kW h = 3.6 xx 106 J.

The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour. For example, electrical energy used during a month is expressed in terms of ‘units’. Here, 1 ‘unit’ means 1 kilowatt hour.

Activity ____________11.17

♦ Take a close look at the electric meter installed in your house. Observe its features closely.

♦ Take the readings of the meter each day at 6.30 am and 6.30 pm.

♦ How many ‘units’ are consumed during day time?

♦ How many ‘units’ are used during night?

♦ Do this activity for about a week.

♦ Draw inferences from the data.

♦ Compare your observations with the details given in the monthly electricity bill.

Q 3215201169

Two girls, each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Class 9 Chapter 11 Example 7
Solution:

(i) Power expended by girl A:
Weight of the girl, m g = 400 N

Displacement (height), h = 8 m
Time taken, t = 20 s
From Eq. (11.8),
Power, P = Work done/time taken

 = (mgh)/t

 = ( 400 N xx 8 m)/(20s)

= 160 W.

(ii) Power expended by girl B:
Weight of the girl, mg = 400 N
Displacement (height), h = 8 m
Time taken, t = 50 s
Power,  P = (m g h)/t

 = ( 400 N xx 8 m)/(50s)

= 64 W.
Power expended by girl A is 160 W.
Power expended by girl B is 64 W.
Q 3215301260

A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 c m, find his power. Take g = 10 m s^(–2).
Class 9 Chapter 11 Example 8
Solution:

Weight of the boy,
m g = 50 k g xx 10 m s^(–2) = 500 N
Height of the staircase,
h = 45 xx 15//100 m = 6.75 m
Time taken to climb, t = 9 s
From Eq. (11.8),
power, P = Work done/time taken

 = (mgh)/t

 = ( 500 N xx 6.75 m)/(9s)

 = 375 W.

Power is 375 W.
Q 3225301261

An electric bulb of 60 W is used for 6 h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Class 9 Chapter 11 Example 9
Solution:

Power of electric bulb = 60 W
= 0.06 kW.
Time used, t = 6 h
Energy = power xx time taken
= 0.06 kW xx 6 h
= 0.36 kW h
= 0.36 ‘units’.
The energy consumed by the bulb is 0.36 ‘units’.