Question
what is the force constant of an ideal spring if 1.02J are stored in the spring when it is compressed 0.22m
Answers
GPT 3.5
To determine the force constant (k) of an ideal spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In mathematical terms, this can be expressed as:
F = -kx
where F is the force applied to the spring, k is the force constant, and x is the displacement of the spring.
Given that 1.02 J of energy are stored in the spring when it is compressed by 0.22 m, we can use the formula for potential energy stored in a spring:
PE = (1/2)kx^2
where PE is the potential energy stored in the spring.
Setting the potential energy equal to 1.02 J and the displacement (x) equal to 0.22 m, we can solve for k:
1.02 J = (1/2)k(0.22 m)^2
1.02 J = (1/2)k(0.0484 m^2)
Simplifying:
1.02 J = 0.0242 k
k = 1.02 J / 0.0242
k ≈ 42.15 J/m
Therefore, the force constant of the ideal spring is approximately 42.15 J/m.
F = -kx
where F is the force applied to the spring, k is the force constant, and x is the displacement of the spring.
Given that 1.02 J of energy are stored in the spring when it is compressed by 0.22 m, we can use the formula for potential energy stored in a spring:
PE = (1/2)kx^2
where PE is the potential energy stored in the spring.
Setting the potential energy equal to 1.02 J and the displacement (x) equal to 0.22 m, we can solve for k:
1.02 J = (1/2)k(0.22 m)^2
1.02 J = (1/2)k(0.0484 m^2)
Simplifying:
1.02 J = 0.0242 k
k = 1.02 J / 0.0242
k ≈ 42.15 J/m
Therefore, the force constant of the ideal spring is approximately 42.15 J/m.