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 # 8 m#, what is the net force on the object?

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Answer 1 · 2018-05-19T14:12:12

#63.2# newtons

The net force of the spring is basically the force exerted on the spring by Hooke's law and the weight of the object.

Hooke's law states that:

#F=kx#

where:

#F# is the force applied in newtons

#k# is the spring constant

#x# is the extension of the spring

So we get:

#F_1=3 \ "kg/s"^2*8 \ "m"#

#=24 \ "kg m/s"^2#

#=24 \ "N"#

The weight of the object is:

#F_2=4 \ "kg"*9.8 \ "m/s"^2#

#=39.2 \ "N"#

So, the total net force on the object is:

#F_"net"=F_1+F_2#

#=24 \ "N"+39.2 \ "N"#

#=63.2 \ "N"#