A force of 0.8N stretches an elastic spring by 5cm.cm find the elastic constant of the spring
The elastic constant of a spring is given by the formula:
k = F/x
where F is the applied force and x is the resulting elongation or compression of the spring.
In this case, F = 0.8 N and x = 5 cm = 0.05 m.
Thus,
k = F/x = 0.8 N / 0.05 m = 16 N/m
Therefore, the elastic constant of the spring is 16 N/m.
If a force of 100N stretches a spring by 0.1cm,find the elastic constant.
The elastic constant of a spring is given by the formula:
k = F/x
where F is the applied force and x is the resulting elongation or compression of the spring.
In this case, F = 100 N and x = 0.1 cm = 0.001 m.
Thus,
k = F/x = 100 N / 0.001 m = 100,000 N/m
Therefore, the elastic constant of the spring is 100,000 N/m.
To find the elastic constant (k) of a spring, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.
Hooke's Law can be expressed as:
F = k * x
Where F is the force applied to the spring, k is the elastic constant (also known as the spring constant), and x is the displacement of the spring.
In this case, we are given the force (0.8N) and the displacement (5cm).
Converting the displacement to meters:
5cm = 0.05m
Using Hooke's Law, we have the equation:
0.8N = k * 0.05m
To find k, we can rearrange the equation:
k = 0.8N / 0.05m
Evaluating the expression:
k = 16 N/m
Therefore, the elastic constant (spring constant) of the spring is 16 N/m.