Question

When a 4.0 kg mass is hung vertically on a light spring that obeys Hooke's law, the string stretches 2.0 cm. How much work must an external agent do to stretch the spring 4.0 cm from its equilibrium position? none of these 1.34J 18.55J 3.14 J 8.55J

Answer 1

According to Hooke's law, the work done to stretch the spring can be calculated using the force and displacement. The correct answer is 1.34 J.

Step-by-step explanation:

Hooke's law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. This relationship can be expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

Here, the displacement of the mass is given as 2.0 cm, and we need to find the work done to stretch the spring by 4.0 cm. The work done on the spring is equal to the area under the force-displacement graph.

If we assume the spring constant is the same, the force can be calculated using Hooke's law, and then multiplied by the displacement to find the work done. The correct answer is 1.34 J.

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