Physics: 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?

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A spring on a horizontal surface can be stretched

taehyungg

02.02.2022

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

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

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

=> $W_{2.5} = 250 \ J$

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