Answer:
F=13.89N
Step-by-step explanation:
A 2.07 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.41 N/m. The mass is displaced 3.15 m to the right from its equilibrium position and then released, which initiates simple harmonic motion.
What is the force (including direction, choose the right to be positive) acting on the mass 3.30 s after it is released?
from hookes law the force applied on an elastic material is directly proportional to the extension, provided that the elastic limit is not exceeded.The general equation for simple harmonic motion along the x-axis is given as thus
k=force constant 4.41N/m
x=is displacement in the x direction 3.15m
F=kx
the mass is displaced 3.15m to the right
F=4.41*3.15m
F=13.89N
take not also that
F=ma
a=d^2x/dt^2
differentiation of speed dx/dt will give acceleration on the body
acceleration is also a=
\
but since we have gotten the force on the object, we have reached the point