Question

A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much work is done in stretching the spring 3.5 m from its equilibrium​ position? b. How much work is done in compressing the spring 2.5 m from its equilibrium​ position?

Answer 1

Answer:

a   $W_{3.5} = 490 \ J$

b  $W_{2.5} = 250 \ J$

Explanation:

Generally the force constant is mathematically represented as

       $k = \frac{F}{x}$

substituting values given in the question

=>   $k = \frac{16}{0.2}$

=>   $k = 80 \ N /m$

Generally the workdone  in stretching the spring 3.5 m is mathematically represented as

       $W_{3.5} = \frac{1}{2} * k * (3.5)^2$

=>     $W_{3.5} = \frac{1}{2} * 80 * (3.5)^2$

=>    $W_{3.5} = 490 \ J$

Generally the workdone  in compressing the spring 2.5 m is mathematically represented as

        $W_{2.5} = \frac{1}{2} * k * (2.5)^2$

=>      $W_{2.5} = \frac{1}{2} * 80 * (2.5)^2$

=>       $W_{2.5} = 250 \ J$