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
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