A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equilibrium length. A 100 g hard plastic ball is then loaded into the tube. If the ball is shot directly up and reaches a height of 2 m above the top of the tube, what is the spring constant, k? Ignore air resistance.

A) 98 N/m
B) 20 N/m
C) 12 N/m
D) 25 N/m
E) 390 N/m

The spring constant is K=392.4N/m

Explanation:

According to hook's law theapplied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

i need the answer to this too but idk it ; -;