Answer:
The ratio is KE : TM = 0.75
Explanation:
from the question we are told that
The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position
Generally the total mechanical energy of the mass is mathematically represented as

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring
Generally this total mechanical energy is mathematically represented as

=> 
Here the potential energy of the mass is mathematically represented as
![PE = \frac{1}{ 2} * k * [ x ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20x%20%5D%5E2)
Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

So
![PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20%5Cfrac%7BA%7D%7B2%7D%20%20%5D%5E2)
So
![KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2](https://tex.z-dn.net/?f=KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20A%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20%5B%5Cfrac%7BA%7D%7B2%7D%20%5D%5E2)
=> 
=> 
So the ratio of
is mathematically represented as

=>
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
It is false. The effect of freezing is almost the exact opposite