Question

When a 2.50 - kg object is hung vertically on a certain light spring described by Hooke’s law, the

spring stretches 2.76 cm. (a) What is the force constant of the spring? (b) If the 2.50 - kg object is

removed, how far will the spring stretch if a 1.25 - kg block is hung on it? (c) How much work must

an external agent do to stretch the same spring 8.00 cm from its unstretched position?​

Answer 1

The work done in the spring is calculated to be 2.8 J

What is Hooke's law?

Hooke's law states that, the extension of a given material is directly proportional to the applied force as long as the elastic limit is not exceeded . First, we must bear in mind that the material must remain within the elastic limit for us to apply the Hooke's law in solving the problem.

Now;

From Hooke's law;

F = Ke

F = force applied

K = force constant

e = extension

F = W = mg =  2.50 - kg * 9.8 m/s^2 = 24.5 N

K = 24.5 N/ 2.76 * 10^-2

K = 888 N/m

e = F/K

F = W =  1.25 - kg * 9.8 m/s^2 = 12.25 N

e = 12.25 N/ 888 N/m = 0.014 m or 1.4 cm

Work done by an external agent = 1/2 Kx^2

= 0.5 * 888 * (8 * 10^-2)^2

= 2.8 J

Learn more about Hoke's law:https://brainly.com/question/13348278

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