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I found this question, and coudn't crack it, and also didn't understand it. Q. A weight of mass m is attached to a vertical spring. The extension in the spring is: a)mg/k b)2mg/k c) mg/2k d)none of these the answer is supposed to be 2mg/k. Can anyone please explain?

I found this question, and coudn't crack it, and also didn't understand it.


Q. A weight of mass m is attached to a vertical spring. The extension in the spring is:


a)mg/k b)2mg/k c) mg/2k d)none of these


the answer is supposed to be 2mg/k. Can anyone please explain?

Grade:

1 Answers

Menka Malguri
39 Points
9 years ago

Initially,the equilibrium position of the vertical spring is at its natural length.

When the mass m is attached to it,it becomes elongated and its equilibrium position reaches a new point.The mass m is attracted by the gravitational force of the earth and the mass m pulls the spring with the same force.The spring applies an equal and opposite restoring force on the mass m and the they remain in rest at equilibrium.

Acccording to this condition,the mass m remains at rest,this means that zero net force acts on it.So,the sum of the forces acting on it must be zero.

Let the elongation of the spring be x,then the restoring force on the mass by the spring is (-kx),where k is the force constant of the spring and negative sign indicates that the direction of the restoring force on the mass by the string is opposite to the direcion of elongation of the spring.And mg is the force acting on the mass due to the gravitational force.And these forces operate in opposite directions.

So,solve further taking help of the concept above.

Hope this works.

Best of Luck!!!!

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