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

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle

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