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Harman [31]
3 years ago
7

A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

ant is 125 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.8 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
Physics
1 answer:
avanturin [10]3 years ago
6 0

Answer:

0.17547 m

Explanation:

m = Mass of block = 3.3\times 10^{-2}\ kg

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m

The amplitude of the resulting simple harmonic motion is 0.17547 m

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Answer:

The time interval is  t = 21.30 \ s

Explanation:

From the question we are told that

    The constant acceleration is \alpha  = 0.029 \ rad / s^2

    The displacement is  \theta  =  6.58 \ rad

     

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    \theta  =  w_i* t  +  \frac{1}{2} *  \alpha  t^2

given that the blade started from rest

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3 years ago
Hi there, it is your average pathetic Junior. Anyways, I need help on 8&9 ASAP, this assignment is beyond overdue but hey wh
zheka24 [161]

Answer:

a) -31.36 m/s

b) 50.176 m

Explanation:

<h2>a) Velocity of the bag</h2>

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

V_{f}=V_{o} +a.t  (1)

Where:

V_{f} is the final velocity of the supply bag

V_{o}=0 is the initial velocity of the supply bag (we know it is zero because we are told <u>it was "dropped", this means it goes to ground in free fall</u>)

a=g=-9.8m/s^{2} is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

t=3.2s is the time

Knowing this, let's solve (1):

V_{f}=0+(-9.8m/s^{2})(3.2s)  (2)

Hence:

V_{f}=-31.36m/s  Note the negative sign is because the direction of the bag is downwards as well.

<h2>b) Final height of the bag</h2>

In this case we will use the following equation:

y=V_{o}t-\frac{1}{2}gt^{2} (3)

Where:

y is the distance the bag has fallen

V_{o}=0 remembering <u>the bag was dropped</u>

g=-9.8m/s^{2} is the acceleration due gravity (downwards)

t=3.2 s is the time

Then:

y=-\frac{1}{2}gt^{2} (3)

y=-\frac{1}{2}(-9.8m/s^{2})(3.2)^{2} (4)

Finally:

y=50.176 m

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