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3 years ago

Jed drops a 18 kg box off of the Eiffel tower. After 6.1 seconds, how fast is the box moving (neglect air resistance.)

Physics

2 answers:

boyakko [2]3 years ago

4 0

Answer:

Speed, v = 60 m/s

Explanation:

Given that,

Mass of the box, m = 18 kg

Time, t = 6.1 seconds

Initially the box is at rest, u = 0

Let v is the final speed of the box i.e. after 6.1 seconds. We can calculate it using first equation of motion as :

$v=u+at$

Here, a = g

$v=gt$

$v=9.8\times 6.1$

v = 59.78 m/s

v = 60 m/s

So, after 6.1 seconds, the box is moving with a speed of 60 m/s. Hence, this is the required solution.

GuDViN [60]3 years ago

3 0

The box fall with acceleration 10m/s2 due to gravity

velocity/ 6.1 = 180
velocity = 1098 m/s

the answer is A.

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a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

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Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

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$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

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$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

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$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

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$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

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$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

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