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2 years ago

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One end of an elastic cord is fastened to a steel beam. A metal weight with a mass of 68 kg is attached to the other end of the

cord and dropped from the level of the beam. The weight is traveling at 14 m/s downward when the rest length of the cord is exceeded and the cord begins to stretch and exert an upward force on the metal weight. After 0. 85 seconds, the weight is moving upward with a speed of 11 m/s. What is the average force on the weight during this time period?.

1 answer:

Vika [28.1K]2 years ago

4 0

Hi there!

We can use IMPULSE to solve.

Recall impulse:

$I = \Delta p = m\Delta v = m(v_f - v_i)$

Begin by calculating the impulse. Assuming up to be the + direction, and down to be the - direction.

$I = 68(11 - (-14)) = 68 \cdot 25 = 1700 kg\frac{m}{s}$

Now, we can calculate force using this value:

$I = F \cdot t\\\\F = \frac{I}{t}\\\\F = \frac{1700}{0.85} = \boxed{2000 N}$

<u>The weight experiences a net force of 2000N UPWARDS.</u>

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Given a point in rectangular form, that is $P(x,y) = (x,y)$ , its polar form is defined by:

$P(x,y) = (r,\theta)$ (1)

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$\theta$ - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

$r = \sqrt{x^{2}+y^{2}}$ (2)

And direction is calculated by following trigonometric relation:

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If we know that $x = -3.50\,m$ and $y = -2.50\,m$ , then the components of coordinates in polar form is:

$r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}$

$r \approx 4.301\,m$

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$\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)$

$\theta \approx 215.538^{\circ}$

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

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