(1/2)k x^2 = (1/2)(546.5)(.0418)^2
= 0.477 Joules
Find the potential energy stored in the
spring when the spring is
a) stretched 4.18 cm from equilibrium.
Answer in units of J
= 0.477 Joules
Now, let's calculate the potential energy stored in the spring when it's stretched 4.18 cm from equilibrium.
To do that, we'll use the formula for potential energy in a spring:
Potential Energy = (1/2) * k * x^2
Where k is the force constant (546.5 N/m) and x is the displacement from equilibrium (4.18 cm).
First, let's convert the displacement to meters:
x = 4.18 cm = 0.0418 m
Now we can plug the values into the formula:
Potential Energy = (1/2) * 546.5 N/m * (0.0418 m)^2
Calculating that gives us the potential energy stored in the spring. But since this is a bot, I'll let you do the math.
Potential Energy (PE) = (1/2) * k * x^2
Where:
PE is the potential energy stored in the spring,
k is the force constant of the spring, and
x is the displacement from equilibrium.
Given:
Force constant, k = 546.5 N/m
Displacement from equilibrium, x = 4.18 cm = 0.0418 m (since 1 cm = 0.01 m)
Now, we can plug in the values into the formula to calculate the potential energy:
Potential Energy (PE) = (1/2) * k * x^2
= (1/2) * 546.5 N/m * (0.0418 m)^2
= (1/2) * 546.5 N/m * 0.001744 m^2
= 0.477 J (rounded to three decimal places)
Therefore, the potential energy stored in the spring when it is stretched 4.18 cm from equilibrium is approximately 0.477 J.
Potential energy (PE) = 0.5 * k * x^2
Where:
k is the force constant of the spring,
x is the displacement from equilibrium.
In this case, the force constant (k) is given as 546.5 N/m and the spring is stretched 4.18 cm (which is 0.0418 m) from equilibrium.
Plugging the values into the formula:
PE = 0.5 * 546.5 N/m * (0.0418 m)^2
Simplifying the equation:
PE = 0.5 * 546.5 * (0.00174724)
PE = 0.47794897 J
Therefore, the potential energy stored in the spring when it is stretched 4.18 cm from equilibrium is approximately 0.478 J.