If the force of 100N stretches a spring by 0.1cm.Find:
1.The elastic constant
2.The work done in stretching the spring by 0.1cm if the elastic limit is not exceeded
To find the elastic constant of a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed.
Hooke's Law is given by:
F = k * x,
where F is the force applied, k is the elastic constant (also known as the spring constant), and x is the displacement.
In this case, a force of 100N stretches the spring by 0.1cm. We can convert 0.1cm to meters:
x = 0.1cm = 0.1 / 100m = 0.001m.
Now we can calculate the elastic constant:
100N = k * 0.001m.
Solving for k, we have:
k = 100N / 0.001m
= 100,000 N/m.
Therefore, the elastic constant of the spring is 100,000 N/m.
To find the work done in stretching the spring by 0.1cm, we can use the formula for work:
W = (1/2) * k * x^2,
where W is the work done, k is the elastic constant, and x is the displacement.
In this case, the displacement x is 0.001m. Plugging in the values, we get:
W = (1/2) * 100,000 N/m * (0.001m)^2
= (1/2) * 100,000 N/m * 0.000001 m^2
= 0.05 J.
Therefore, the work done in stretching the spring by 0.1cm is 0.05 Joules.