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4 years ago

7

Coil springs are widely used in industry. Those used in scientific instruments must be manufactured to exhibit consistent perfor

mance. One important property of a spring is its force constant. The quality control manager of a coil spring manufacturer is examining a coil spring made by her company that compresses 0.0510 m0.0510 m when a mass of 175 kg175 kg is placed on it. Calculate the force constant ????.

1 answer:

Strike441 [17]4 years ago

5 0

Answer:

33627.4 N/m

Explanation:

By Hook's law F = k x

where F= force applied

K= Spring constant

x= distance it compresses.

Since the force applied in this case will be the weight of the object that causes it to compress, so, the weight will be (175 * 9.8) = 1715N force which causes a compression (x) of 0.0510m.

So,

F = K x

K = F / x

K = 1715 / 0.0510

K = 33627N/m

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A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m

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Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

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Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

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Part B

The conservation of energy will be used again.

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}$

where $k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89$

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Part C

Total energy of the system is equal to the potential energy at amplitude.

$E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J$

Part D

The general equation of motion in simple harmonic motion is

$x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)$

where $\phi$ is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

$x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

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