An object with a mass of #4 kg# is hanging from a spring with a constant of #3 (kg)/s^2#. If the spring is stretched by # 6 m#, what is the net force on the object?

Answer 1

#F_n=21,24 " N"#

#G=m*g# #G=4*9,81=39,24 N" Object's weight"# #F=-k*Delta x# #F=-3*6=-18 " N Force acting on spring"# #F_n=G+F# #F_n=39,24-18# #F_n=21,24 " N"#
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Answer 2

The net force acting on the object can be calculated using Hooke's Law:

[ F = kx ]

where:

  • ( F ) is the force,
  • ( k ) is the spring constant, given as ( 3 ) (kg)/s(^2),
  • ( x ) is the displacement, which is ( 6 ) m.

Plugging in the values:

[ F = (3)(6) ]

[ F = 18 ]

Therefore, the net force acting on the object is ( 18 ) Newtons.

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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.

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.

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.

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.

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