Question
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.
Answers
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.
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.