if a force of 100 newton stretches a spring by 5 meters .calculate the
1)elastic constant
2)workdone
Where is the answer to my question
1) k = force / stretch = 100 N / 5 m = 20 N/m
2) work = 1/2 k x^2 = 1/2 * 20 * 5^2 ... Joules
To calculate the elastic constant and the work done, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it stretches or compresses.
1) Elastic Constant (k):
We can use Hooke's Law to find the elastic constant (k) of the spring. Hooke's Law can be written as F = k * x, where F is the force applied, k is the elastic constant, and x is the displacement.
Given:
Force (F) = 100 N
Displacement (x) = 5 m
Rearranging Hooke's Law to solve for the elastic constant (k):
k = F / x
k = 100 N / 5 m
k = 20 N/m
Therefore, the elastic constant (k) of the spring is 20 N/m.
2) Work Done (W):
The work done on a spring is given by the equation W = (1/2) * k * x^2, where W is the work done, k is the elastic constant, and x is the displacement.
Given:
Elastic constant (k) = 20 N/m
Displacement (x) = 5 m
Substituting the given values into the equation:
W = (1/2) * 20 N/m * (5 m)^2
W = (1/2) * 20 N/m * 25 m^2
W = 10 N/m * 25 m^2
W = 250 N
Therefore, the work done on the spring is 250 joules.