(a) 2446 N/m
When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:
![E=U=\frac{1}{2}kA^2](https://tex.z-dn.net/?f=E%3DU%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2)
where
U is the elastic potential energy
k is the spring constant
A is the maximum displacement (the amplitude)
Here we have
U = E = 50.9 J
A = 0.204 m
Substituting and solving the formula for k,
![k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2E%7D%7BA%5E2%7D%3D%5Cfrac%7B2%2850.9%29%7D%7B%280.204%29%5E2%7D%3D2446%20N%2Fm)
(b) 50.9 J
The total mechanical energy of the system at any time during the motion is given by:
E = K + U
where
K is the kinetic energy
U is the elastic potential energy
We know that the total mechanical energy is constant: E = 50.9 J
We also know that at the equilibrium point, the elastic potential energy is zero:
because x (the displacement) is zero
Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:
![K=E=50.9 J](https://tex.z-dn.net/?f=K%3DE%3D50.9%20J)
(c) 8.55 kg
The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when
K = 50.9 J (at the equilibrium position)
Kinetic energy can be written as
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where m is the mass
Solving the equation for m, we find the mass:
![m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B2K%7D%7Bv%5E2%7D%3D%5Cfrac%7B2%2850.9%29%7D%7B%283.45%29%5E2%7D%3D8.55%20kg)
(d) 2.14 m/s
When the displacement is
x = 0.160 m
The elastic potential energy is
![U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7D%282446%29%280.160%29%5E2%3D31.3%20J)
So the kinetic energy is
![K=E-U=50.9 J-31.3 J=19.6 J](https://tex.z-dn.net/?f=K%3DE-U%3D50.9%20J-31.3%20J%3D19.6%20J)
And so we can find the speed through the formula of the kinetic energy:
![K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%5Crightarrow%20v%3D%5Csqrt%7B%5Cfrac%7B2K%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2819.6%29%7D%7B8.55%7D%7D%3D2.14%20m%2Fs)
(e) 19.6 J
The elastic potential energy when the displacement is x = 0.160 m is given by
![U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7D%282446%29%280.160%29%5E2%3D31.3%20J)
And since the total mechanical energy E is constant:
E = 50.9 J
the kinetic energy of the block at this point is
![K=E-U=50.9 J-31.3 J=19.6 J](https://tex.z-dn.net/?f=K%3DE-U%3D50.9%20J-31.3%20J%3D19.6%20J)
(f) 31.3 J
The elastic potential energy stored in the spring at any time is
![U=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where
k = 2446 N/m is the spring constant
x is the displacement
Substituting
x = 0.160 m
we find the elastic potential energy:
![U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7D%282446%29%280.160%29%5E2%3D31.3%20J)
(g) x = 0
The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.