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KATRIN_1 [288]
3 years ago
8

A vertical spring stretches 4.4 cm when a 16-g object is hung from it. the object is replaced with a block of mass 22 g that osc

illates up and down in simple harmonic motion. calculate the period of motion.
Physics
1 answer:
Yuri [45]3 years ago
7 0
Using K= mg/x
where k is a unction of the construction of the spring and has nothing to do with the mass attached to it;
k = 0.016 × 9.81 / 0.044
   = 3.57 N/M
when  a mass of 22 g is used the system will oscillate with a period T = 2π √(m/k)
= 2 × 3.14 ×√ 0.025/3.57)
= 6.28 × 0.0837
= 0.5255
Therefore, the period will be 0.5255 seconds
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To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef
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29.76245 rad/s², -117.80972 rad/s²

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\omega_f = Final angular velocity

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\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2

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\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2

Angular acceleration during spin down is -117.80972 rad/s²

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\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad

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