A spring of force constant 120N/m is acted upon by a constant force of 240N.Calculate the elastic potential energy stored in the spring. Please with full solution
Solution:
The elastic potential energy stored in the spring is given by the equation:
U = 1/2 * k * x^2
where k is the force constant and x is the displacement of the spring.
In this case, k = 120N/m and x can be calculated using the equation:
F = k * x
x = F/k
x = 240N/120N/m
x = 2m
Therefore, the elastic potential energy stored in the spring is given by:
U = 1/2 * 120N/m * (2m)^2
U = 480J
Thank you
To calculate the elastic potential energy stored in the spring, we can use the formula:
Elastic potential energy = (1/2) * k * x^2
where:
- k is the force constant of the spring
- x is the extension or compression of the spring from its equilibrium position
In this case, the force constant (k) is given as 120 N/m and the force (F) acting on the spring is 240 N.
The force acting on the spring is related to the extension or compression through Hooke's Law:
F = k * x
Rearranging the equation, we can solve for x:
x = F / k
Substituting the given values:
x = 240 N / 120 N/m
= 2 m
Now, we can calculate the elastic potential energy:
Elastic potential energy = (1/2) * k * x^2
= (1/2) * 120 N/m * (2 m)^2
= (1/2) * 120 N/m * 4 m^2
= 240 J
Therefore, the elastic potential energy stored in the spring is 240 J.
To calculate the elastic potential energy stored in a spring, you can use the formula:
Elastic Potential Energy (PE) = 1/2 * k * x^2
where k is the force constant of the spring and x is the displacement of the spring from its equilibrium position.
In this case, the force constant (k) is given as 120 N/m, and the spring is acted upon by a constant force of 240 N. To find the displacement (x), we can use Hooke's Law:
F = k * x
Rearrange the equation to solve for x:
x = F / k
Substituting the given values:
x = 240 N / 120 N/m
x = 2 m
Now we can calculate the elastic potential energy:
PE = 1/2 * k * x^2
PE = 1/2 * 120 N/m * (2 m)^2
PE = 1/2 * 120 N/m * 4 m^2
PE = 1/2 * 480 Nm
PE = 240 Nm
Therefore, the elastic potential energy stored in the spring is 240 Nm.