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

Solution:

\[ \displaystyle \Large  F=40N, s=5m\]

$F = 40N, s = 5m$

Therefore,

\[ \displaystyle \Large W=Fs\]

\[ \displaystyle \Large W=(40)(5)\]

\[ \displaystyle \Large W=200J\]

$W = Fs$

$W = (40)(5)$

Arthur will need to do \(200J\) of work on the box to it at least \(5m\).

Next

$F = 30\sqrt{2} N$

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 force applied is at an angle. Therefore we need to find the horizontal force first. 

\[ \displaystyle \Large F= 60 \times \cos(45)\]

\[ \displaystyle \Large F=30 \sqrt{2} N\]

$F = 60*cos(45)$

Now we can use the normal work equation.

\[ \displaystyle \Large W=Fs\]

\[ \displaystyle \Large W= (30 \sqrt{2}) (8)\]

\[ \displaystyle \Large W= 240 \sqrt{2} J= 340 J\]

$W = Fs$



Next

A weightlifter lifts a \(100 kg\) barbell from the ground to a height of 2 meters in 4 seconds. Calculate the power exerted by the weightlifter during this lifting process.

Solution:

In order to calculate the power we must find the work done by the weightlifter.

\[ \displaystyle \Large W= Fs= (100 \times 9.81)(2)= 1962 J \]

$W = Fs = (100 * 9.81)(2) = 1962J$

We are given that the time is \(4s\). Therefore we have everything to use the power formula now.

\[ \displaystyle \Large P= \frac{W}{t}=\frac{1962}{4}=490.5 W \]

$P = \frac{W}{t}=\frac{1962}{4}=490.5W$

Next