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:
[tex]E=PE+KE=const.[/tex]
The potential energy is given by
[tex]PE=\frac{1}{2}kx^2[/tex]
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:
[tex]E=KE_{max}=\frac{1}{2}mv_{max}^2[/tex] (1)
where
m is the mass of the block
[tex]v_{max}=1.25 m/s[/tex] is the maximum speed
We also know that the total energy is
[tex]E=0.18 J[/tex]
Re-arranging eq.(1), we can find the mass:
[tex]m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg[/tex]
b)
The maximum speed in a spring-mass system is also given by
[tex]v_{max} =\sqrt{\frac{k}{m}} A[/tex]
where
k is the spring constant
m is the mass
A is the amplitude
Here we have:
[tex]v_{max}=1.25 m/s[/tex] 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
[tex]k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m[/tex]
c)
The frequency in a spring-mass system is given by
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
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:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz[/tex]
d)
We can solve this part by using the law of conservation of energy; in fact, we have
[tex]E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2[/tex]
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:
[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s[/tex]
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