A spring has a force constant of 60000 N/m.
How far must it be stretched for its potential energy to be 47 J?
Answer in units of m
PE = 1/2 kx^2
To calculate the distance the spring must be stretched for its potential energy to be 47 J, we can use the formula for potential energy of a spring:
U = 1/2 * k * x^2
where U is the potential energy, k is the force constant, and x is the distance stretched. Rearranging the formula, we have:
x = sqrt(2 * U / k)
Substituting the given values:
x = sqrt(2 * 47 J / 60000 N/m)
Calculating:
x = sqrt(0.001567)
x ≈ 0.0396 m
Therefore, the spring must be stretched approximately 0.0396 meters for its potential energy to be 47 J.
To find the distance the spring must be stretched, we can use Hooke's Law, which states that the potential energy of a spring is given by the equation:
PE = (1/2) * k * x^2
where PE is the potential energy, k is the force constant of the spring, and x is the distance the spring is stretched or compressed.
In this case, we are given the force constant (k = 60000 N/m) and the potential energy (PE = 47 J), and we need to solve for x.
Rearranging the equation, we have:
47 J = (1/2) * 60000 N/m * x^2
Multiplying both sides by 2 to cancel out the (1/2), we get:
94 J = 60000 N/m * x^2
Dividing both sides by 60000 N/m, we have:
x^2 = 94 J / 60000 N/m
x^2 = 0.001567
To solve for x, we take the square root of both sides:
x = sqrt(0.001567)
Calculating this, we find that x is approximately 0.0396 m.
Therefore, the spring must be stretched by approximately 0.0396 meters for its potential energy to be 47 J.