An object with a mass of #3 kg# is hanging from a spring with a constant of #8 (kg)/s^2#. If the spring is stretched by #2 m#, what is the net force on the object?

Question

1545 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-16T09:03:23

#13.43# #"N"# downwards

For this problem, let's consider upwards to be the positive direction.

Gravity is acting downwards on the object, i.e. in the negative direction. So #g# will be equal to #- 9.81# #"m s"^(- 2)#.

Also, the spring is stretched by #2# #"m"# (also downwards). So it will be equal to a displacement of #- 2# #"m"#.

Now, the net force will be the sum of the upwards and downwards forces acting on the object.

Let's use #F = ma# and #F = - k x#:

#Rightarrow F_("net") = ma + (- k x)#

#Rightarrow F_("net") = 3# #"kg" cdot (- 9.81)# #"m s"^(- 2) - 8# #"kg s"^(- 2) cdot (- 2)# #"m"#

#Rightarrow F_("net") = - 29.43# #"kg m s"^(- 2) + 16# #"kg m s"^(- 2)#

#Rightarrow F_("net") = - 13.43# #"kg m s"^(- 2)#

#therefore F_("net") = - 13.43# #"N"#

Therefore, the net force acting on the object is #13.43# #"N"# downwards.