Question

A force of 5n is applied to an elastic spring of length 2m makes it to extend by 0.01m what will be the new length if a force of 10n is applied

Answer 1

Given that a force of 5N extended the spring by 0.01m when the original length was 2m, we can calculate the spring constant k using Hooke's Law:

F = kx

5N = k(0.01m)

k = 500 N/m

Now, when a force of 10N is applied, we can find the new extension by rearranging Hooke's Law:

F = kx

10N = 500N/m * x

x = 0.02m

Therefore, the new length of the spring when a force of 10N is applied will be 2.02m.