A certain spring stores 35.0 J of potential energy when it is compressed by 9.2 cm. What is the spring constant?
KE=1/2 k x^2
k=2*KE/x^2=70/.092^2 N/m
To find the spring constant, we can use the formula for potential energy stored in a compressed spring:
Potential energy (U) = (1/2) * k * x²
where:
U = potential energy
k = spring constant
x = displacement from equilibrium position
Given:
U = 35 J (potential energy)
x = 9.2 cm = 0.092 m (displacement)
Substituting the given values into the formula:
35 J = (1/2) * k * (0.092 m)²
Simplifying the equation:
70 J = k * (0.008464 m²)
To find k, we can divide both sides of the equation by (0.008464 m²):
k = 70 J / (0.008464 m²)
Calculating the value of k:
k ≈ 8269.9 N/m
Therefore, the spring constant is approximately 8269.9 N/m.
To find the spring constant, we can use Hooke's Law, which states that the potential energy stored in a spring is directly proportional to the square of its displacement (extension or compression) from its equilibrium position. The equation for potential energy of a spring is given by:
PE = (1/2)kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement.
In this case, the potential energy (PE) is given as 35.0 J and the displacement (x) is given as 9.2 cm. However, we need to convert the displacement to meters in order to use SI units consistently:
1 meter = 100 centimeters
So, 9.2 cm = 9.2/100 = 0.092 m
Now we can substitute these values into the equation and solve for the spring constant (k):
35.0 J = (1/2)k(0.092 m)^2
35.0 J = (1/2)k(0.008464 m^2)
Now, we can solve for k by rearranging the equation:
k = (35.0 J) / [(1/2)(0.008464 m^2)]
k = (35.0 J) / (0.004232 m^2)
k ≈ 8262.39 N/m
So, the spring constant is approximately 8262.39 N/m.