An object with a mass of #6# #kg# is hanging from a spring with a constant of #12# #kgs^-2#. If the spring is stretched by # 2# #m#, what is the net force on the object?

Answer 1

Hooke's Law says: #F=kx# - the force acting is the product of the spring constant and the displacement of the object from the spring's equilibrium position:

#F=kx=12xx2=24# #N#

(To answer this specific question, we do not need to know the object's mass.)

I have mentioned this in other answers, but I prefer the unit #Nm^-1# rather than #kgs^-2# for expressing the spring constant.

The two units are equivalent, since one newton is one #kgms^-2#.

Given that we are usually talking about forces in #N# and displacements in #m#, the unit #Nm^-1# is just more directly useful.

Just keep in mind that they both mean the same thing and are directly equivalent.

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Answer 2

Adam Arnold

The net force on the object is 24 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.