Question

What is the normal force acting on a 70kg person riding a drop tower when they are falling

at an acceleration of 9.8 m/s^2

Answer 1

686 N

Step-by-step explanation:

OH BOY, NEWTONS LAWS OF MOTION!

look at this:

`F = ma`

a fun fact is that acceleration 9.8 m/s^2 is actually GRAVITY.

so lets do this and rewrite the formula...

`F = 70kg * 9.8 m/s^2`

lets multiply 70 and 9.8

we get 686 N.

if you want to be technical about it...... the answer should only have two sig figs. so it would be...
`6.9 * 10^(2)` N

NICE!