Question

a force of 5 Newton extends on a spring by 2.5 CM calculate the spring constant of the spring and work done by the force​

Answer 1

Answer:

12.5 probably will be the answer

Explanation:

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Answer 2

Given:

A force of 5 N extends a spring from 2 cm to 5 cm.

$F = 5 N, x_1 = 2 cm, x_2 = 5 cm$

To find:

(i) The spring constant of the spring

(ii) The work done by the force

Solution:

(i) The spring constant of the spring

The formula to compute the spring constant is,

$k = -\dfrac{F}{x}\\\Rightarrow k = - \dfrac{5}{0.02-0.05}\\ \therefore k = 167 N/m$

(ii) The work done by the force

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\Rightarrow W=\dfrac{1}{2} \times 167 \times (0.05^2-0.02^2)\\\Rightarrow W=0.175 N/m\\\therefore W=0.175 J$

Therefore,

(i) The spring constant of the spring is, 167 N/m

(ii) The work done by the force is, 175 mJ