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Ksenya-84 [330]
3 years ago
13

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?
Physics
1 answer:
nekit [7.7K]3 years ago
5 0

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

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