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Ivenika [448]
3 years ago
14

1. Find the magnitude of the gravitational force a 66.5 kg person would experience

Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

Gravitational forces: Earth: W = 650.969\,N, Mars: W = 246.449\,N, Pluto: W = 38.504\,N

Explanation:

The weight (W), measured in newtons, experimented by the person on the surface of the planet is defined by Newton's Laws of Motion:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

From Newton's Law of Gravitation we derive this expression for gravitational acceleration:

g = \frac{G\cdot M}{R^{2}} (2)

Where:

G - Gravitational constant, measured in cubic meters per kilogram-square second.

M - Mass of the planet, measured in kilograms.

R - Radius of the planet, measured in meters.

By (2) we calculate the gravitational acceleration for each planet:

Earth (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.97\times 10^{24}\,kg, R = 6.38\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.97\times 10^{24}\,kg)}{(6.38\times 10^{6}\,m)^{2}}

g = 9.789\,\frac{m}{s^{2}}

Mars (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 6.42\times 10^{23}\,kg, R = 3.40\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (6.42\times 10^{23}\,kg)}{(3.40\times 10^{6}\,m)^{2}}

g = 3.706\,\frac{m}{s^{2}}

Pluto (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 1.25\times 10^{22}\,kg, R = 1.20\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1.25\times 10^{22}\,kg)}{(1.20\times 10^{6}\,m)^{2}}

g = 0.579\,\frac{m}{s^{2}}

Lastly, we calculate the gravitational force for each case by (1):

Earth (m = 66.5\,kg, g = 9.789\,\frac{m}{s^{2}})

W =(66.5\,kg)\cdot \left(9.789\,\frac{m}{s^{2}} \right)

W = 650.969\,N

Mars (m = 66.5\,kg, g = 3.706\,\frac{m}{s^{2}})

W = (66.5\,kg)\cdot \left(3.706\,\frac{m}{s^{2}} \right)

W = 246.449\,N

Pluto (m = 66.5\,kg, g = 0.579\,\frac{m}{s^{2}})

W = (66.5\,kg)\cdot \left(0.579\,\frac{m}{s^{2}} \right)

W = 38.504\,N

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Answer:

a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s

Explanation:

a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?

We write the equation of the forces acting on the mass.

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So, T - mg = ma

T/m - g = a

dv/dt = T/m - g

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dv/dt = (10.3 m/s²)t - 9.8 m/s²

dv = [(10.3 m/s²)t - 9.8 m/s²]dt

Integrating, we have

∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt

∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt

v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C

v = (5.15 m/s³)t² - (9.8 m/s²)t + C

when t = 0, v = 0 (since at t = 0, box is at rest)

So,

0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C

0 = 0 + 0 + C

C = 0

So, v = (5.15 m/s³)t² - (9.8 m/s²)t

i. What is the velocity of the box at t = 1.00 s,

v =  (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)

v = 5.15 m/s - 9.8 m/s

v = -4.65 m/s

ii. What is the velocity of the box at t = 3.00 s,

v =  (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)

v = 15.45 m/s - 29.4 m/s

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b. What is the maximum distance that the box descends below its initial position?

Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t

dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt

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∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt

∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt

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when t = 0, y = 0.

So,

0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'

0 = 0 + 0 + C'

C' = 0

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

The maximum distance is obtained at the time when v = dy/dt = 0.

So,

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0

(5.15 m/s³)t² - (9.8 m/s²)t = 0

t[(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or (5.15 m/s³)t = (9.8 m/s²)

t = 0 or t = (9.8 m/s²)/(5.15 m/s³)

t = 0 or t = 1.9 s

Substituting t = 1.9 s into y, we have

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²

y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)

y = 11.798 m - 17.689 m

y = -5.891 m

y ≅ - 5.89 m

So, the maximum distance that the box descends below its initial position is 5.89 m

c. At what value of t does the box return to its initial position?

The box returns to its original position when y = 0. So

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

0 = (1.72 m/s³)t³ - (4.9 m/s²)t²

(1.72 m/s³)t³ - (4.9 m/s²)t² = 0

t²[(1.72 m/s³)t - (4.9 m/s²)] = 0

t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0

t = √0 or (1.72 m/s³)t = (4.9 m/s²)

t = 0 or t = (4.9 m/s²)/(1.72 m/s³)

t = 0 or t = 2.85 s

So, the box returns to its original position when t = 2.85 s

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