Explanation:
the question is unanswerable
r = radius of the circle of the ride = 3.00 meters
v = linear speed of the person during the ride = 17.0 m/s
m = mass of the person in angular motion in the ride
L = angular momentum of the person in the ride = 3570 kg m²/s
Angular momentum is given as
L = m v r
inserting the values
3570 kg m²/s = m (17 m/s) (3.00 m)
m = 3570 kg m²/s/(51 m²/s)
m = 7 kg
hence the mass comes out to be 7 kg
For the answer to the question above, this is the maximum displacement, the spring has only elastic potential energy.
spring is constant @ 5 N/m
maximum displacement = 2 cm = 0.02 m
elastic potential energy = 1/2 kx²
= 0.5 x 5 x 0.02²
So the answer would be
= 0.001 Joules
So 1 kg = 2.2 pounds.
66kg | 2.2 pounds
--------| ------------------
| 1kg
I set it up like this. The 66 kg crosses out with the 1kg. So you multiply the top 66 x 2.2 = 145.2 pounds
There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.
