Power word problems
Below are some carefully selected power word problems to help you understand the concept of power in physics.
Problem # 1:
How much power is required to do 200 joules
of work in 4 seconds?What if you do the same work in 2 seconds? Did you
notice anything? Say what you noticed.
Solution:
Power =
work done
/
time interval
Power = 50 Watts or 50 joules per second
What if the same work is done in 2 seconds?
Power = 100 Watts or 100 joules per second
I noticed that more power is required to do the same work in less time.
Interesting power word problems
Problem # 2:
Which person is more powerful?
It takes John 2 seconds to lift a 100-kg barbell a distance of 20 centimeters.
It takes Peter 3 seconds to lift a 200-kg barbell a distance of 15 centimeters.
Solution:
First, convert 20 centimeters and 15 centimeters to meters.
20 centimeters = 0.20 meter and 15 centimeters = 0.15 meter.
Find the force that is required to lift 100 kg and 150 against gravity.
F = 100 × 10 = 1000 Newtons
F = 200 × 10 = 2000 Newtons
Power =
work done
/
time interval
Power =
F × d
/
time interval
John's power =
1000 × 0.2
/
2
John's power =
1000
/
2
× 0.2
Power = 500 × 0.2
Power = 100 Watts or 100 joules per second
John's power =
2000 × 0.15
/
3
Power = 100 Watts or 100 joules per second
John and Peter have the same power.
Problem # 3:
The power of a machine is 4 kilowatts. How
much work can this machine do in 1 second ? Answer should be in Newtons/
4 meters. Don't use the formula to solve this problem!
If you still cannot do it without using the formula,
click here to see the solution.
Solution:
4 kilowatts = 4000 watts
Remember that 1 watt = 1 joule of work in 1 second
4000 watts then = 4000 joules of work in 1 second
The work done then in 1 second = 4000 joules.
Remember that 1 joule means 1 Newton of force over a distance of 1 meter.
4000 joules means 4000 Newtons of force over a distance of 1 meter.
This means 16000 Newtons of force over a distance of 4 meters
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