A spring 20cm is stretched to 25cm by load of 50n. What will be it's length when stretched by 100n. Assuming that the elastic limit is not reached. 40cm,10cm,35cm,40cm,50cm.
We can use Hooke's law to solve this problem:
F = kx
where F is the force applied, x is the displacement from the equilibrium length, and k is the spring constant.
We can first find the spring constant using the initial data:
k = F/x = 50 N / (25 cm - 20 cm) = 10 N/cm
Now we can use this spring constant to find the displacement at 100 N:
x = F/k = 100 N / 10 N/cm = 10 cm
Therefore, the length of the spring when stretched by 100 N will be:
25 cm + 10 cm = 35 cm
So the answer is option C, 35 cm.
To find the length of the spring when stretched by 100N, we can use Hooke's Law, which states that the force exerted on a spring is directly proportional to the extension of the spring.
Hooke's Law can be mathematically represented as:
F = k * x
Where:
F = force applied to the spring (N)
k = spring constant (N/m)
x = extension of the spring (m)
We can rearrange the equation to solve for the extension:
x = F / k
Given that the spring is initially stretched from 20cm to 25cm by a load of 50N, we can determine the spring constant:
x = ΔL = 25cm - 20cm = 0.05m
F = 50N
x = F / k
0.05m = 50N / k
Solving for k:
k = 50N / 0.05m
k = 1000 N/m
Now, we can use this spring constant to find the length of the spring when stretched by 100N:
F = 100N
k = 1000 N/m
x = F / k
x = 100N / 1000 N/m
x = 0.1m = 10cm
Therefore, the length of the spring when stretched by 100N is 10cm.