A force of a magnitude 500N is applied to a Free end of a spiral spring of contant 1.0*10NM,Cal the energy stored in a streched spring
if you mean k = 1.0*10 N/m, then
F = kx
E = 1/2 kx^2 = 1/2 Fx
Not sure why you would specify k = 1.0*10 instead of just saying k=10. Did you leave off an exponent?
Yes
To calculate the energy stored in a stretched spring, you can use the formula:
E = (1/2) k x^2
Where:
E is the energy stored in the spring
k is the spring constant
x is the displacement from the equilibrium position
Given:
Force (F) = 500 N
Spring constant (k) = 1.0 * 10 N/m
To calculate the displacement (x), we need to rearrange the formula for the spring constant:
k = F / x
Rearranging the above equation, we get:
x = F / k
Substituting the given values:
x = 500 N / (1.0 * 10 N/m)
x = 50 m
Now, we can calculate the energy stored in the spring:
E = (1/2) * k * x^2
Substituting the values:
E = (1/2) * (1.0 * 10 N/m) * (50 m)^2
E = (1/2) * 1.0 * 10 N * (2500 m^2)
E = 1.0 * 10 J * 2500
E = 25000 J
Therefore, the energy stored in the stretched spring is 25000 Joules.
To calculate the energy stored in a stretched spring, we need to use the formula:
E = (1/2)kx²
Where:
E is the energy stored in the spring
k is the spring constant
x is the displacement of the spring from its equilibrium position
In this case, the magnitude of the force applied to the spring is 500N, and the spring constant is given as 1.0*10 N/m. However, the displacement (x) is not provided in the question. Therefore, we need to first determine the displacement of the spring.
To find the displacement, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:
F = kx
Rearranging the equation, we have:
x = F/k
Plugging in the values, we get:
x = 500 N / (1.0*10 N/m) = 50 m
Now that we have the displacement, we can proceed to calculate the energy stored in the spring:
E = (1/2)kx²
Plugging in the values, we get:
E = (1/2)(1.0*10 N/m)(50 m)² = 12500 J
Therefore, the energy stored in the stretched spring is 12500 Joules.