# Lesson 20: Power

**Notice**: Trying to get property 'post_author' of non-object in

**/home/forge/cilearning.ca/web/app/themes/buddyboss-theme/learndash/ld30/lesson.php**on line

**168**

Download here: Ontario Curriculum Expectations

As knowledge increases, wonder deepens.

Charles Morgan

#### Practice Questions

- Fred and Barney have a job moving large rocks at the Bedrock quarry. Mr. Slate decides to have a contest to see which of his employees can do more work. Eager to win Fred picks up a 10.0 kg rock and runs up a hill as fast as he can raising the rock to a height of 20.0 m in only 20.0 s before falling over exhausted. Barney, working a little slower, manages the same feat but takes a full minute to accomplish the task. Who has done more work?

**Answer**

FRED:

Given:

m = 10.0 kg

h = 20.0 m

Î”t = 20.0 s

g = 9.8 N/kg

Required:

F_{g} = ? (In order to find work)

W = ?

Solution:

F_{g} = mg

F_{g} =(10 kg)(9.8 N/kg)

F_{g} =98 N

W = (98 N)(20.0 m)

W = 1960 J

Fred has done 1960 J of work.

BARNEY:

Given:

m = 10.0 kg

h = 20.0 m

Î”t = 1 minute = 60.0 s

g = 9.8 N/kg

Required:

F_{g} = ? (In order to find work)

W = ?

Solution:

F_{g} = mg

F_{g} =(10 kg)(9.8 N/kg)

F_{g} = 98 N

W = (98 N)(20.0 m)

W = 1960 J

Barney has done 1960 J of work.

Barney and Fred have both done the same amount of work, 1960 J.

In the above problem, both Fred and Barney have done the same amount of work. However, Fred has done the work much faster than Barney. In physics we need a way to differentiate between the two very different efforts put forth by Fred and Barney to complete the work of lifting the rock to a height of 20.0 m. This can be done by looking at their power.

#### Definition of Power:

Power is:

- the rate at which work is done.
- how fast a force does work.

Since energy is consumed when work is done, power is also the rate at which energy is used.

#### Equation for Power:

Substituting the equation for work (W = FÎ”d) in the above equation yields: .

Since , therefore,

#### Units of Power:

By the working with the power equation, the following can be found.

In physics, J/s is equal to a unit called the Watt, named after James Watt. The symbol for the Watt is the capital letter W. Be careful not to confuse this with the unit for work.

A force develops 1 W of power when it does 1 J of work in 1 s.

### Example

What is the power of a bulldozer that does 5.5 x 104 J of work in 1.1 s?

Given:

W = 5.5 x 104 J

Î”t = 1.1 s

Required:

P = ?

Solution:

The bulldozer does 50 000 W of power.

Now return to the Fred and Barney problem at the beginning of this activity. They have both done the same amount of work, 1960 J, but who is more powerful?

Check out this video to learn more about work and power:

#### Practice Questions

Calculate how much power Fred and Barney each had while lifting the rock in the problem at the beginning of this activity.

**Answer**

FRED: Given: W = 1960 J Î”t = 20.0 s Required: P = ? Solution: Fred has a power of 98 W. | BARNEY: Given: W = 1960 J Î”t = 60.0 s Required: P = ? Solution: Barney has a power of 32.7 W. |

Suzie and Amanda have a bet to see who is the most powerful. Suzie applies a force of 200 N to push a box 20.0 m in 40.0 s. Amanda applies the same force and is able to move the box at a velocity of 0.75 m/s. Who has the greater power output?

**Answer**

SUZIE: Given: F = 200 N Î”d = 20.0 m Î”t = 40.0 s Required: W = ? (to solve for Power) P = ? Solution: W = FÎ”d W = (200 N)(20.0 m) W = 4000 J Suzie has a power output of 100 W. | AMANDA: Given: F = 200 N v = 0.75 m/s Required: P = ? Solution: P = Fv _{avg}P = (200 N)(0.75 m/s) P = 150 W Amanda has a power output of 150 W. |

As can be seen in the calculations, Amanda has a greater power output.

A car on a level road accelerates from 0 m/s to 28.0 m/s with an average acceleration of 5.0 m/s^{2}. The total friction on the car is 600 N. Calculate the power of the engine if the car has a mass of 1400 kg.

**Answer**

Given:

0 m/s

28 m/s

5.0 m/s^{2}

-600 N (This value is negative because it opposes the motion of the car.)

m =1400 kg

Required:

? (to find )

? (to find power)

Î”t = ? (to find power)

Solution:

ma =(1400 kg)(5.0 m/s^{2}) = 7000 N

= 7000 â€“ (-600 N)

=7600 N

P = Fv_{avg}

P = (7600 N)(14 m/s)

P = 106 400 W

The power of the car engine is about 106 kW.