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 =  
200 / 4

Power = 50 Watts or 50 joules per second

What if the same work is done in 2 seconds?

Power =  
200 / 2

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

John's power =  
300 / 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.

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