a spring 20cm long is stretched to 25cm by a load of 50N. what will be its lenght when stretched by 100N assuming that the elastic limit is not reached

ANSWER

k = 50N/(25-20)cm = 10N/cm.

Length = 20cm + 100N*1cm/10N = 30 cm.

30CM

To determine the length of the spring when stretched by 100N, we can use Hooke's Law. Hooke's Law states that the force exerted on a spring is directly proportional to the extension or compression of the spring, as long as the elastic limit is not reached.

First, let's calculate the spring constant, denoted by k. The spring constant represents the stiffness of the spring. It can be found using Hooke's Law equation: F = k * x, where F is the force applied to the spring and x is the extension or compression of the spring.

Given that the spring is initially 20cm long and is stretched to 25cm by a load of 50N:
x1 = 25cm - 20cm = 5cm (extension)
F1 = 50N

We can rearrange the equation F = k * x to solve for k:
k = F / x = 50N / 5cm = 10N/cm

Now, let's find the extension of the spring when a load of 100N is applied:
F2 = 100N
k = 10N/cm

Using the equation F = k * x, we can rearrange it to solve for x:
x2 = F2 / k

Plugging in the values, we have:
x2 = 100N / 10N/cm = 10cm

Therefore, when stretched by 100N, the length of the spring will be 10cm longer than its original length of 20cm, resulting in a length of 30cm.