a spring 20cm long is streched to 25cm by a load of 50n what will be it length when streched by 100n assuming that the elastic limit is not reached show soving
K=F/e
K=50/25-20
K=10N/cm
e=F/K
e=100/10
e=10
Therefore new length =20 + 10 = 30cm
The elongation is directly proportional to force.
50n gives 5cm elongation
100N gives 10cm, or a length of 30 cm
To find the length of the spring when stretched by 100 N, we can use Hooke's law, which states that the force applied to a spring is directly proportional to the extension produced.
Hooke's Law: F = k * x
Where:
F is the force applied to the spring (in Newtons)
k is the spring constant (a measure of the stiffness of the spring)
x is the extension or change in length (in meters)
First, we need to find the spring constant, k. We can use the given information that the spring is stretched by 5 cm (from 20 cm to 25 cm) by a load of 50 N.
To find k, we rearrange Hooke's law: k = F / x.
k = 50 N / 0.05 m
k = 1000 N/m
Now that we have k, we can use it to find the change in length (extension) of the spring when a 100 N load is applied.
Using Hooke's law: F = k * x
We rearrange the formula to solve for x: x = F / k.
x = 100 N / 1000 N/m
x = 0.1 m or 10 cm
Therefore, when a load of 100 N is applied, the spring will stretch by 10 cm from its original length of 20 cm. So, the final length of the spring will be 30 cm.