a) The total energy of the system is sum of kinetic energy and elastic potential energy:
[tex]E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2 [/tex]
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring
The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
[tex]x=A=4.00 cm = 0.04 m[/tex]
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
[tex]E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J [/tex]
b) When the position of the object is
[tex]x=1.00 cm = 0.01 m[/tex]
the potential energy of the system is
[tex]U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J [/tex]
and so the kinetic energy is
[tex]K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J[/tex]
since the mass is [tex]m=50.0 g=0.05 kg[/tex], and the kinetic energy is given by
[tex]K= \frac{1}{2}mv^2 [/tex]
we can re-arrange the formula to find the speed of the object:
[tex]v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s [/tex]
c) The potential energy when the object is at
[tex]x=3.00 cm=0.03 m[/tex]
is
[tex]U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J [/tex]
Therefore the kinetic energy is
[tex]K=E-U=0.028 J-0.016 J = 0.012 J[/tex]
d) We already found the potential energy at point c, and it is given by
[tex]U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J [/tex]
