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vampirchik [111]
2 years ago
11

If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?

Physics
1 answer:
never [62]2 years ago
3 0

The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

Explanation:

When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.

spring constant is k = 1.00 * 10^3 N/m

mass is x = 20.0 cm

According to Hooke's law, To find restoring force,

    F = - kx

        = - 1.00 *10 ^3 * 20.0

     F = 20000 N/m

Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

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1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

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If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

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The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

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\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

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\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

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t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

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t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

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