1) The spring constant is 233.2 N/m
2) The oscillation frequency is 0.93 Hz
3) The speed of the block is 3.2 m/s
4) The maximum acceleration is [tex]23.9 m/s^2[/tex]
5) The net force on the block is 104.9 N
Explanation:
1)
At equilibrium, the weight of the mass is equal to the restoring force of the spring. Therefore, we can write:
[tex]mg=kx[/tex]
where
m = 6.9 kg is the mass hanging on the spring
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x = 0.29 m is the stretching of the spring
Solving for k, we find
[tex]k=\frac{mg}{x}=\frac{(6.9)(9.8)}{0.29}=233.2 N/m[/tex]
2)
The oscillation frequency of 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,
k = 233.2 N/m is the spring constant
m = 6.9 kg is the mass
Substituting, we find the frequency:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{233.2}{6.9}}=0.93 Hz[/tex]
3)
The velocity of the block at time t is given by the equation:
[tex]v(t) = -v_0 cos(\omega t)[/tex]
where
[tex]v_0 = 4.1 m/s[/tex] is the initial speed
[tex]\omega[/tex] is the angular frequency
t is the time
The angular frequency can be found from the frequency:
[tex]\omega=2\pi f=2\pi(0.93)=5.84 rad/s[/tex]
And substituting t = 0.42 s, we find the velocity of the block at this time:
[tex]v(t)=-4.1 cos ((5.84)(0.42))=3.2 m/s[/tex]
4)
The maximum acceleration of the block is given by
[tex]a_{max} = \omega^2 A[/tex] (1)
where
[tex]\omega=5.84 rad/s[/tex] is the angular frequency
A is the amplitude
The amplitude is related to the initial velocity by the equation:
[tex]v_0=\omega A[/tex] (2)
Combining (1) and (2), we find
[tex]a_{max}=v_0 \omega[/tex]
And substituting [tex]v_0 = 4.1 m/s[/tex], we find
[tex]a_{max}=(4.1)(5.84)=23.9 m/s^2[/tex]
5)
The acceleration at time t can be found by calculating the derivative of v(t), and it is given by the equation
[tex]a(t) = a_{max} sin(\omega t)[/tex]
where
[tex]a_{max}=23.9 m/s^2[/tex] is the maximum acceleration
[tex]\omega=5.84 rad/s[/tex] is the angular frequency
t is the time
Substituting t = 0.42 s,
[tex]a(t)=(23.9)sin((5.84)(0.42))=15.2 m/s^2[/tex]
Finally, the net force on the block can be found by using Newton's second law:
[tex]F=ma=(6.9)(15.2)=104.9 N[/tex]
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