The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
Answer:
The answer is 3.111111.
Explanation:
It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.
Answer:
Same weight so neither is heavier
