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GarryVolchara [31]
3 years ago
10

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

motion that occurs has a maximum speed of 1.7 m/s. Determine the amplitude A of the motion.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

\displaystyle w=\sqrt{\frac{k}{m}}

The equation for the motion of the object is written as a sinusoid:

\displaystyle X=A\ cos\ w\ t

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

\displaystyle X'=V=-A\ w\ sin\ w\ t

And the maximum speed is

\displaystyle V_{max}= A\ w

Solving for the amplitude

\displaystyle A= \frac{V_{max}}{w}

Computing w

\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\   rad/ s

Calculating A

\displaystyle A=\frac{1.7}{24.5}=0.069\ m

\displaystyle \boxed{A=6.9\ cm}

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Answer:

h = 2.821\,m

Explanation:

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(83\,kg)\cdot(9.807\,\frac{m}{s})\cdot (3.90\,m) = \frac{1}{2}\cdot (83\,kg)\cdot v^{2}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (3.90\,m)}

v \approx 8.746\,\frac{m}{s}

The speed after the inelastic collision is obtained by using the Principle of Momentum Conservation:

(83\,kg)\cdot (8.746\,\frac{m}{s} )+(55\,kg)\cdot (0\,\frac{m}{s} ) = (83\,kg + 55\,kg)\cdot v

v = 5.260\,\frac{m}{s}

Lastly, the maximum height is determined by using the Principle of Energy Conservation again:

\frac{1}{2}\cdot (138\,kg)\cdot (5.260\,\frac{m}{s} )^{2} = (138\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot h

h = \frac{(5.260\,\frac{m}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

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I divided the 200m by 4s (so 200/4) to get 50m per second
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A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?
Semenov [28]

Answer:

The maximum range R_{max}= 132. 72 m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

                                R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }  m

Where,

                          g - acceleration due to gravity

Substituting the values in the above equation

                   R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }  

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Hence, the car lands at a distance, R_{max}= 132. 72 m            

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3 years ago
A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p
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Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

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Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

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Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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