a) The mass is 0.23 kg
b) The spring constant is 1.25 N/m
c) The frequency is 1.42 Hz
d) The speed of the block is 1.08 m/s
Explanation:
a)
We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:
![E=PE+KE=const.](https://tex.z-dn.net/?f=E%3DPE%2BKE%3Dconst.)
The potential energy is given by
![PE=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:
(1)
where
m is the mass of the block
is the maximum speed
We also know that the total energy is
![E=0.18 J](https://tex.z-dn.net/?f=E%3D0.18%20J)
Re-arranging eq.(1), we can find the mass:
![m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B2E%7D%7Bv_%7Bmax%7D%5E2%7D%3D%5Cfrac%7B2%280.18%29%7D%7B%281.25%29%5E2%7D%3D0.23%20kg)
b)
The maximum speed in a spring-mass system is also given by
![v_{max} =\sqrt{\frac{k}{m}} A](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%20A)
where
k is the spring constant
m is the mass
A is the amplitude
Here we have:
is the maximum speed
m = 0.23 kg is the mass
A = 14.0 cm = 0.14 m is the amplitude
Solving for k, we find the spring constant
![k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bv_%7Bmax%7D%5E2%7D%7BA%5E2%7Dm%3D%5Cfrac%7B1.25%5E2%7D%7B0.14%5E2%7D%280.23%29%3D18.3%20N%2Fm)
c)
The frequency in a spring-mass system is given by
![f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
where
k is the spring constant
m is the mass
In this problem, we have:
k = 18.3 N/m is the spring constant (found in part b)
m = 0.23 kg is the mass (found in part a)
Substituting and solving for f, we find the frequency of the system:
![f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7B18.3%7D%7B0.23%7D%7D%3D1.42%20Hz)
d)
We can solve this part by using the law of conservation of energy; in fact, we have
![E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=E%3DPE%2BKE%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Where v is the speed of the system when the displacement is equal to x.
We know that the total energy of the system is
E = 0.18 J
Also we know that
k = 18.3 N/m is the spring constant
m = 0.23 kg is the mass
Substituting
x = 7.00 cm = 0.07 m
We can solve the equation to find the corresponding speed v:
![v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2E-kx%5E2%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%280.18%29-%2818.3%29%280.07%29%5E2%7D%7B0.23%7D%7D%3D1.08%20m%2Fs)
#LearnwithBrainly