Question

A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m.

Answer 1

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 $23.9 m/s^2$

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:

$mg=kx$

where

m = 6.9 kg is the mass hanging on the spring

$g=9.8 m/s^2$ 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

$k=\frac{mg}{x}=\frac{(6.9)(9.8)}{0.29}=233.2 N/m$

2)

The oscillation frequency of a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

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:

$f=\frac{1}{2\pi}\sqrt{\frac{233.2}{6.9}}=0.93 Hz$

3)

The velocity of the block at time t is given by the equation:

$v(t) = -v_0 cos(\omega t)$

where

$v_0 = 4.1 m/s$ is the initial speed

$\omega$ is the angular frequency

t is the time

The angular frequency can be found from the frequency:

$\omega=2\pi f=2\pi(0.93)=5.84 rad/s$

And substituting t = 0.42 s, we find the velocity of the block at this time:

$v(t)=-4.1 cos ((5.84)(0.42))=3.2 m/s$

4)

The maximum acceleration of the block is given by

$a_{max} = \omega^2 A$ (1)

where

$\omega=5.84 rad/s$ is the angular frequency

A is the amplitude

The amplitude is related to the initial velocity by the equation:

$v_0=\omega A$ (2)

Combining (1) and (2), we find

$a_{max}=v_0 \omega$

And substituting $v_0 = 4.1 m/s$, we find

$a_{max}=(4.1)(5.84)=23.9 m/s^2$

5)

The acceleration at time t can be found by calculating the derivative of v(t), and it is given by the equation

$a(t) = a_{max} sin(\omega t)$

where

$a_{max}=23.9 m/s^2$ is the maximum acceleration

$\omega=5.84 rad/s$ is the angular frequency

t is the time

Substituting t = 0.42 s,

$a(t)=(23.9)sin((5.84)(0.42))=15.2 m/s^2$

Finally, the net force on the block can be found by using Newton's second law:

$F=ma=(6.9)(15.2)=104.9 N$

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