U3AOS1 Topic 13: Work and Power

Work is literally what the word means. It's how much energy has been used on the object by a force and is calculated by multiplying the force applied that is parallel to the movement with the distance it occurred over:

\[ \displaystyle \Huge W= Fs\]



Work is done on or by someone or something. If you push a box then the work is done by you because you are losing energy and the work is done on the box as it gains energy.



Created with GeoGebra®, by Tan Seng Kwang, Link


Power is how much work was done over how much time. Work can be from any form of energy (electricity, mechanical, etc). 

\[ \displaystyle \Huge P=\frac{\Delta E}{\Delta t}=\frac{W}{t}\]


A small amount of work over a very small time can create large amounts of power. An example of this is Bruce Lee's one inch punch.




Example 1

Arthur applies $ 40 $ N of force to a box that he is moving. He needs to move the box at least $ 5 $ m. What is the minimum amount of work he will need to do on the box to move it at least $ 5 $ m?

Solution:
Using the values:
\[ F = 40 \text{N} , s = 5 \text{m} \]

We can sub them in the work equation $ W = Fs $ :
\[ W = (40)(5) = 200 \text{J} \]

Arthur will need to do $ 200 $ J of work on the box to move it at least $ 5 $ m.



Example 2

A person pushes a sled with a force of $ 60 $ Newtons at an angle of $ 45 $ degrees to the horizontal. If the sled is displaced horizontally by $ 8 $ meters, calculate the work done.


Solution:
The net force is applied at an angle but the motion is only horizontal. Therefore we need to find the part of the force that's causing the motion, the horizontal component, first:
\[ F_{x} = 60 \cos{45 \degree} \]
\[ F_{x} = 30 \sqrt{2} \text{N} \]

Now that we have the component of the force that's doing the work, we can sub this into the work equation:
\[ W = Fs \]
\[ W = (30 \sqrt{2} \text{N})(8) \]
\[ W = 240 \sqrt{2} \text{J} \approx 340 \text{J} \]


Example 3

A weightlifter lifts a $ 100 $ kg barbell $ 2 $ off the ground. If it takes them $ 4 $ seconds to do this, calculate the power of this lifting.

Solution:

In order to calculate the power, we must find the work done by the weightlifter. Given that the lifting is done against gravity, the force will be $ F_{g} $. We can sub this and the displacement into the work equation to find:
\[ W = Fs = (100 * 9.81) (2) = 1962 \text{J} \]

We are given that the time is $ 4 $ s. Therefore we have everything to use the power formula now:
\[ P = \frac{W}{t} = \frac{1962}{4} = 490.6 \text{W} \]

Exercise 1

Exercise 2

Exercise 3

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