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?

Question

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?

Charles Aeschliman · Accepted Answer

#63.2# newtons

The force that Hooke's law and the object's weight apply to the spring together constitute the net force of the spring. According to Hooke's law, #F=kx# where #F# is the force applied in newtons #k# is the spring constant #x# is the extension of the spring Thus, we obtain: #F_1=3 \ "kg/s"^2*8 \ "m"# #=24 \ "kg m/s"^2# #=24 \ "N"# The object's weight is: #F_2=4 \ "kg"*9.8 \ "m/s"^2# #=39.2 \ "N"# Thus, the object's total net force is: #F_"net"=F_1+F_2# #=24 \ "N"+39.2 \ "N"# #=63.2 \ "N"#

Benjamin Asiedu · Answer

undefined The net force on the object can be found using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.

Plugging in the given values, we have F = -(3 kg/s^2) * 8 m = -24 N.

The negative sign indicates that the force is acting in the opposite direction of the displacement. Therefore, the net force on the object is 24 N.