An object with a mass of #6 kg# is hanging from a spring with a constant of #3 (kg)/s^2#. If the spring is stretched by #9 m#, what is the net force on the object?

Answer 1

The net force is #F_"net"=27 N#.

My interpretation: The mass is initially hanging, in equilibrium, so #F_"net"=0#. Then an outside force pulls it down 9m from the initial position. I could say that the mass is now in another equilibrium and then say again that #F_"net"=0#. But that would be trivial.

As a result, I believe that when the external force is abruptly released, we are asked to provide the net force.

Hooke's Law gives us that the spring's additional force due to that 9m displacement is #F = k*x = 3 (kg)/s^2 * 9 m = 27 (kg*m)/s^2 = 27 N# This is the net force on the mass when freed at that position.

After the external force is eliminated, the mass will begin to bounce, and the net force will change depending on where it is.

Hope this is helpful, Steve.

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Answer 2

Lily Adams

The net force on the object is 18 N.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.