Question

A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm?

Answer 1

Answer:36 J

Explanation:

(2400+1200)÷2 ×0.02= 36 J

Answer 2

Answer:

Work done performed is -24joule (This work is done by external applied force not by the spring).

Explanation:

Provided that:

F = 1200 N force is applied to compress a spring from its natural length of 18 cm to 16 cm .

So, Initial length (L1) = 18 cm

Final length (L2) = 16 cm

From Hooke's law we know that:

F = -K (L1 - L2) ; where K is Spring Constant.

$F = -K (L1 - L2) \: ...equation(i)$

From equation (i) we can calculate the value of K for the given spring;

$F = -K (L2 - L1) \\ or \: K = \frac{F}{(L1 - L2)} \\ or \: K = \frac{1200}{(18 - 16)} \: newton. {cm}^{ - 1} \\ = 600 \: newton. {cm}^{ - 1} \\ = 60000 \: newton. {m}^{ - 1}$

Now to compress the spring from 16 cm to 14 cm;

external force needed;

$F = -K (L2 - L1) \\ F = 600 \times (16 - 14) \: newton \\ = 1200 \: newton$

Work done to compress the spring from 16 cm to 14 cm;

$W = F \: . \: ds \\ or \: W=F \: . (L2 - L1) \\ W= - 1200 \times 0.02 \: joule \\ = - 24 \: joule$

So the Work done performed is -24 joule .

Negetive sign (-ve) indicates this work done is made by external force not by the system (Spring). And this work done is stored in the spring as Potential energy.