TERM: 2nd TERM
SUBJECT: BASIC SCIENCE
CLASS: JSS 2
WEEK THREE DATE:………………..
TOPIC: CALCULATION INVOLVING WORK DONE
Worked Example 1
How much work is done when a force of 5 kN moves its point of application 600mm in the direction of the force.
W.D = (5×103)×(600 × 10-3)
Worked Example 2
Find the work done in raising 100 kg of water through a vertical distance of 3m.
The force is the weight of the water, so
W.D = (100 × 9.81) × 3
Work done by a variable force
Worked example 3
What is the potential energy of a 10kg mass: 100m above the surface of the earth at the bottom of a vertical mine shaft 1000m deep.
= 10 × 9.81 × 100
= 10 × 9.81 × (-100)
Worked example 4
A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a constant breaking force over a distance of 75m.
= 500 ×302
= 500 × 102
The SI unit for power is the watt W.
A power of 1W means that work is being done at the rate of 1J/s.
Larger units for power are the kilowatt kW (1kW = 1000 W = 103 W) and the megawatt MW (1 MW = 1000000 W = 106 W).
If work is being done by a machine moving at speed v against a constant force, or resistance, F, then since work doe is force times distance, work done per second is Fv, which is the same as power.
Ques I; If a force of one Newton is applied by a car over a distance of 5metres , what is the work done?
Work done =Force x distance, therefore
W= f x d
200 x 6= 1200joules
Ques 2: Calculate the power of a pump which can lift 500kg of water through a vertical height of 12m in 0.3 minutes ,assuming g =9.8m/s
Force =500kg x 9.8N
Work done =500 x 9,8 x 12j
Time taken = 0.3 x 60 = 18 sec
Power = Work done = 500 x 9.8 x 12 j
Time taken 18
Ques 3: A student whose mass is 60kg runs up a height of 7.2 m in 10.4 second . Find the power used by the student (given that g = 9.8m/s) .
Force = 60kg x 9.8N
Work done = F x D
= 60 x 9.8 x 7.2 j
Power == Work done = 60 x 9.8 x 7.2
Time taken 10.4sec
Precious seed BASIC SCIENCE FOR JUNIOR SECONDARY SCHOOLS BOOK 2 PAGE 117
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