A spring 20 cm long, is stretched to 25 cm by a load of 50N.What will be it's length when stretched by 100N assuming that the elastic limit is not reached F=ke

x=force/k

so if you increse force by 2x, stretch will be 2x, or 5cm*2=10cm

what is the answer

please I n exeed the answer

To solve this problem, we need to use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring as long as the elastic limit is not reached. Mathematically, it is expressed as F = kx, where F is the force applied, k is the spring constant, and x is the displacement (change in length) of the spring.

Given that the spring is initially 20 cm long and stretched to 25 cm by a load of 50N, we can find the spring constant, k.

Using Hooke's Law: F = kx
50N = k(25 cm - 20 cm)
50N = k(5 cm)

Simplifying the equation:
k = 50N / 5 cm
k = 10 N/cm

Now, we can use the spring constant to find the length of the spring when stretched by 100N.

Using Hooke's Law: F = kx
100N = (10 N/cm) * x

Simplifying the equation and solving for x:
x = 100N / 10 N/cm
x = 10 cm

Therefore, the length of the spring when stretched by 100N will be 10 cm longer than its original length of 20 cm. Hence, it will be 30 cm long.