Respuesta :
1) The spring constant is 272 N/m
2) The oscillation frequency is 0.96 Hz
3) The speed of the block at t = 0.33 s is 1.7 m/s
4) The maximum acceleration is [tex]25.3 m/s^2[/tex]
5) The net force on the block at t = 0.33 s is 173.3 N
6) The potential energy is maximum at the highest point of the oscillation
Explanation:
1)
When the block is hanging on the spring and the spring is in equilibrium, it means that the weight of the block is balancing the restoring force of the spring. Therefore, we can write:
[tex]mg=kx[/tex]
where
[tex]mg[/tex] is the weight of the block, with
m = 7.5 kg being the mass
[tex]g=9.8m/s^2[/tex] is the acceleration of gravity
[tex]kx[/tex] is the restoring force in the spring, with
[tex]k[/tex] being the spring constant
x = 0.27 m is the stretching in the spring
Solving for k, we find the spring constant:
[tex]k=\frac{mg}{x}=\frac{(7.5)(9.8)}{0.27}=272 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
Here we have
k = 272 N/m is the spring constant
m = 7.5 kg is the mass
Substituting, we find the frequency of oscillation:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{272}{7.5}}=0.96 Hz[/tex]
3)
The block is pushed downward with speed
[tex]v_0 = 4.2 m/s[/tex]
at [tex]t=0[/tex]. This means that we can write the equation of its speed at time t as
[tex]v(t) = -v_0 cos(\omega t)[/tex]
where
[tex]\omega = 2\pi f = 2\pi(0.96)=6.02 rad/s[/tex] is the angular frequency of the system
We can verify that for [tex]t=0[/tex], the equation gives [tex]v(0)=-v_0[/tex], so it gives the correct initial velocity (the negative sign means downward)
Now we can substitute t = 0.33 s to find the velocity of the block at that time:
[tex]v(0.33)=-(4.2)(cos (6.02\cdot 0.33))=+1.7 m/s[/tex]
So the speed is 1.7 s.
4)
The maximum acceleration in a spring-mass system is given by
[tex]a_{max}=\omega^2 A[/tex] (1)
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude
To find the maximum acceleration, we can use also the following relationship:
[tex]v_0 = \omega A[/tex] (2)
Divinding (1) by (2),
[tex]\frac{a_{max}}{v_0}=\omega[/tex]
And so we find
[tex]a_{max}= v_0 \omega = (4.2)(6.02)=25.3 m/s^2[/tex]
5)
The acceleration of the block at time t is equal to the derivative of the velocity, therefore
[tex]a(t) = \omega^2 A sin(\omega t) = a_{max} sin(\omega t)[/tex]
where
[tex]a_{max}=25.3 m/s^2[/tex] is the maximum acceleration of the system
[tex]\omega=6.02 rad/s[/tex] is the angular frequency
By substituting t = 0.33 s, we find the acceleration:
[tex]a(0.33)=(25.3)sin(6.02\cdot 0.33)=23.1 m/s^2[/tex]
And therefore, the net force on the block is equal to the product between its mass and its acceleration:
[tex]F=ma=(7.5)(23.1)=173.3 N[/tex]
6)
The potential energy of the system is given by the sum of the gravitational potential energy ([tex]E_g[/tex]) and the elastic potential energy ([tex]E_e[/tex]):
[tex]E=E_g+E_e[/tex]
where
[tex]E_g = mgh[/tex], where
m is the mass of the block
g is the gravitational acceleration
h is the height of the block
[tex]E_e=\frac{1}{2}k(x'-x)^2[/tex], where
k is the spring constant
(x'-x) is the new stretching of the spring relative to the equilibrium position (x=0.27 m)
We notice that:
- [tex]E_e[/tex] is maximum when [tex](x-x')[/tex] is maximum, so at the highest point of the oscillation and at the lowest point of the oscillation
- [tex]E_g[/tex] is maximum when [tex]h[/tex] is maximum, so at the highest point of the oscillation
Therefore, if we add the two contribution, the potential energy is maximum when both [tex]U_g, U_e[/tex] are maximum, so at the highest point of the oscillation.
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