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3 years ago

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A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

d–Earth system relative to the child's lowest position at the following times. (a) when the ropes are horizontal J.(b) when the ropes make a 34.0° angle with the vertical J.(c) when the child is at the bottom of the circular arc J.

1 answer:

o-na [289]3 years ago

4 0

Answer:

a) U = 735 J , b) U = 125.7 J , c) U = 0 J

Explanation:

The gravitational power energy is

U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

y = 2.10 m

w = mg = 350 N

U = 350 2.10

U = 735 J

b) when the angle is 34º

y = L - L cos 34

y = L (1- cos34)

y = 2.10 (1- cos 34)

y = 0.359 m

U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

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Explanation:

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

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$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

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Region II (t = 4 s to t = 6 s)

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$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

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Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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