A string 20cm long is stretch 25cm by a load of 50n . what will be it's length when stretched by 100n assuming that the elastic limit is not reached

Answer 1

To solve this problem, we need to use Hooke's Law, which states that the extension of a spring is directly proportional to the load applied, as long as the elastic limit is not reached. In other words:

F=kx

where F is the load applied, x is the extension of the spring, and k is the spring constant. The spring constant is a measure of the stiffness of the spring and is usually given in units of N/m.

To find the spring constant, we can rearrange the equation as:

k=F/x

In this case, we know that the spring is 20cm long and is stretched by 25cm when a load of 50N is applied. Therefore:

x=25cm-20cm=5cm=0.05m

F=50N

k=50N/0.05m=1000N/m

Now we can use Hooke's Law again to find the extension of the spring when a load of 100N is applied:

F=100N

k=1000N/m

x=F/k

x=100N/1000N/m=0.1m

Therefore, the spring will be stretched by 0.1m, or 10cm, when a load of 100N is applied. Its total length will be:

20cm+10cm=30cm