A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm

0.25 J

Explanation:

Given that

Force on the spring, F = 30 N

Natural length of the spring, l1 = 20 cm = 0.2 m

Final length of the spring, l2 = 35 cm = 0.35 m

Extension of the spring, x = 0.35 - 0.2 = 0.15 m

The other extension is 40 - 35 cm = 5 cm = 0.05 m

Work done = ?

Considering Hooke's Law of Elasticity.

We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm

The question is solved in the attachment below

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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