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An object with a mass of #5"kg"# is hanging from a spring with a constant of #3 "kg/s"^2#. If the spring is stretched by #3"m"#, what is the net force on the object?

Answer 1

Emily Abbate

#41 "N"# downwards.

The object is subject to a gravitational force that is

#W = mg = (5 "kg") * (10 "m/s"^2) = 50 "N"#

downward.

By Hooke's Law, the spring force is determined.

#F_"Spring" = kx = (3 "kg/s"^2) * (3 "m") = 9 "N"#

upwards.

As a result, the total of all forces acting on the object is the net force.

#F_"Net" = W - F_"Spring"#

#= 50 "N" - 9 "N"#

#= 41 "N"#

downward.

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Answer 2

Carson Alexander

The net force on the object is ( F = kx = (3 , \text{kg/s}^2)(3 , \text{m}) = 9 , \text{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.