MET IS 480 Physics: Module 4 Assignment on Gravitation and Energy

This physics assignment, addressing Module 4, delves into the principles of gravitation and mechanical energy. It begins by analyzing Bob Beamon's record-breaking long jump, discussing the physics factors at play, including gravity, altitude, and air resistance in Mexico City. The assignment then explores gravitational force calculations between objects, comparing the force between cars to Earth's gravity, and determines astronaut's weight on different planets. Further questions cover orbital periods of exoplanets, stopping distances for vehicles, and comparisons of kinetic energy in different scenarios. The assignment also involves calculations related to roller coasters, work, potential energy, and the advantages of using ramps. The document provides detailed solutions, calculations, and explanations for each problem, drawing upon concepts from Newton's laws and principles of energy.

Running Head: Module 4 1
Module 4 Assignment
Student’s Name
University Affiliation
Date

Module 4 2
Module 4 Assignment
1. One of the greatest feats in history of summer Olympics was Bob Beamon’s world record
long jump in 1968in Mexico City. It is fitting that Bob Beamon was one of the first
individuals to inducted into the USA Olympics Hall of Fame; his jump was an
extraordinary accomplishment. Nonetheless, were there any physics factors that aided
Bob Beamon that day in Mexico City to break the prior worlds record by almost two
feet? Discuss two of these two factors. Briefly explain how these factors influence
Beamon’s performance? Look at the table below and also consider the characteristics of
Mexico City.

Module 4 3
Answer
Factors which influenced Beamon’s performance
Position of Mexico City on earth surface
Mexico City lies within the equator which has a lesser force gravity (centrifugal
force cancels out the gravity minimally) as compared to the poles which experience more
gravitational force due to it being closer to the center of the earth as result of bulging of
the earth in the equator hence experience greater gravitational fields.
Altitude of Mexico City
It is located at a higher altitude which experience less gravitational force due
increase in distance from the center of gravity.
Air resistance
Mexico City had an altitude of 2239 m (7342 ft) hence had a
thinner air which means it had a lesser air resistance than it would
have had a lower altitude like the sea level. There was also less drag
from the wind, Beamon benefitted from a trailing wing of 2m/s which is
the highest allowed wind speed for this game.
2. What is the gravitational force between two 1000 kg cars separated by a distance of 50
meters on interstate highway? What is the weight of these cars? Compare the magnitude
of these two forces.
Answer
Gravitational force between the two bodies is:
F= Gm1m2

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r2
where G is the universal gravitational constant
m1- 1000 kg
m2- 1000 kg
r- 50 meters
G- 6.67408 × 10-11 m3 kg-1 s-2
Solution;
F= Gm1m2
r2
F= 6.67408 × 10-11 m3 kg-1 s-2 * 1000 kg * 1000 kg
(50m )2
F= 2.669632 * 10-08 kgms-2 (Newton)
What is the weight of these cars?
Weight of each car will be
Weight= mass * gravitational force
Weight= 1000 kg * 9.8 m/s2
Weight=9800 Newton
Total cars weight= 2 * 1000 kg * 9.8 m/s2= 19600 kgms-2 (Newton)
Compare the magnitude of these two forces.
The gravitational acceleration due to the earth’s gravity has a greater magnitude
as compared to the gravitational force between the cars, this is due to the extremely
massive mass of the earth compared to the masses of the two cars.

Module 4 5
3. The acceleration of gravity on Jupiter is 26.7 m/s2. If an astronaut weighs 160 lbs. on
Earth, what is the weight on Jupiter?
Answer
Weight = Mass * Acceleration
Calculations:
Mass of the astronaut= 160lbs = 72.5747792 kg
Weight on Jupiter= 72.5747792 kg * 26.7 m/s2= 1937.7466 Newton
4. An astronaut’s mass is 50 kg, and thus she weighs about 110 lbs. on Earth. What is her
weight on a small planet with 1/80 the mass of the earth and 1/3 the width of the earth?
Answer
g= Gm1m2; G= 6.67408 × 10-11 m3 kg-1 s-2
r2
where:
g- is gravitational acceleration
m- planets mass
r- is the radius of the planet
Earths mass= 5.973 * 2024 kg
Smalls planets mass= 5.973 * 2024 kg ÷ 80 = 7.467 * 1022 kg
Earths radius= 6378 km
Smalls planets= 6378000 m ÷ 3 = 2126000 m
g of the small planet= 6.6742 * 10-11 * (7.467 * 1022 ÷ 2126000)
g of the small planet= 1.1025m/s2
weight= mass * gravitational force (g)

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An astronaut’s weight on the small planet = 50kg * 1.1025 m/s2 = 55.125 Newton
5. An exoplanet, a plane revolving around a star other than the sun, was discovered recently
by a Boston University Scientist. The mass of the star has been estimated as 1.4 * 1031kg,
which is several times more massive than our sun, and the average distance between the
planet and the star has been measured as 2.3 * 109 kilometers, how long does the planet
take to orbit once around the star?
Answer
Solution
r3 = GM
T2 4 π2
T2= (r3 * 4 π2) ÷ (G * M)
T2= ((2.3 * 1012m)3 * 4 π2) ÷ (6.67408 × 10-11 m3 kg-1 s-2 * 1.4 * 1031kg)
T2= 4.98614766×1045
T= 7.06126593×1022 seconds
6. When Harold drives his pickup truck on US Route 1 in Maine at 35km/hour and brakes
for pedestrians, he can stop in a distance of 9.0 meters. One day when Harold was driving
on an interstate highway at 105km/hour, a moose strolls onto the highway 100 meters
ahead of him. If Harold decelerates his truck at the same rate as on US Route 1. Will he
be able to stop before hitting the moose? What distance will Harold need to stop his
truck?
Answer
Step 1:
V2= u2 + 2as

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Where:
v-final speed
u-initial speed
a-deceleration
s- distance covered decelerating
Calculations
0 = (9.722m/s)2 + 2* a * 9m
0= 94.52 +18a
-94.52m2/s2 = 18m a
-94.52 /18 m/s2 =a
-5.25m/s2=a
Deceleration= -5.25m/s2
Step 2:
Using the same formulae
V2= u2 + 2as
Where:
v-final speed
u-initial speed
a-deceleration
s- distance covered decelerating
To find stoppage distance, then,
0 m/s = (29.17 m/s)2 + 2 * -5.25m/s2 *S
S= -850.8889 m2/s2 ÷ 10.5 m/s2

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S = 81.0370 m
Harold will need a stoppage distance of, S= 81.0370 m
Yes, Harold will stop the truck at 81.0370 m before hitting the moose which is at 100 m
away.
7. Which will have a greater kinetic energy when it lands: a ball dropped from 5 meters that
lands on the ground or another ball with the identical mass that is dropped from 10 meters
above the bottom of a 20-meter well? Explain.
Answer
KE= mgh,
The ball which is dropped at 10 meters above the bottom of the well will have a
greater kinetic energy.
Explanation
Assuming both the balls are 1 kg each because they are identical, then;
Ball on the land will fall a height of 5 meters on the ground while the ball inside the well
will fall a height of 10 meters.
Calculations:
Ball on land: KE= 1kg * 9.8m/s2* 5m = 49 kg m2 / s2
Ball on well: KE= 1kg * 9.8m/s2* 10m = 98 kg m2 / s2
8. A roller coaster car of mass 1000kg starts from rest at a height of 40 meters. Assume
there is no friction in the system.
9. What will be the speed of the car when it reaches the bottom of the of the roller coaster
ramp?
Answer

Module 4 9
Step 1: Find the time of travel
S= ut + 0.5 at2
S- distance
u- initial speed
t – travel time
a- gravitational acceleration
Calculation:
40 m= 0 * t + 0.5 * 9.8 m/s2 *t2
40 m= 0 + 4.9 m/s2 *t2
t2= 40/4.9 m/s2
t2= 8.1633 s2
t2= 2.8525 seconds
Step 2: Find the final speed
V= u + at
v- final speed
u- initial speed
a- gravitational acceleration
t- time taken
Calculations:
v= 0 + 9.8 m/s2 * 2.8525 secs.
V= 27.9545 m/s

Module 4 10
Final speed / speed at the bottom of the roller coaster = 27.9545 m/s
a) Does the answer in part (a) depend on the mass of the car?
Answer.
No. The answer depends only on the vertical height
b) Do the people on the roller coaster car scream as it goes down the ramp?
Answer.
Yes, because of that exhilarating feeling of free fall in such a span of 2.8525 seconds
that comes with the earth’s gravity acceleration.
10. Suppose you wish to compare a work done by pushing a box up on rollers up the ramp to
the work done if you lift the straight up the same final height.
a. What work will be required to lift a 178 N box (about 140lbs) up to a table which
is 0.8m off the floor?
Solution
Work = force * distance
Work = 178 N * 0.8 m
Work = 142.4 Joules
b. Les assume you have a ramp available that makes an angle of 30o with the
horizontal. The ramp is 1.6 m long. How much work does it require to push the
box up the ramp, assuming there is no friction?
Solution
Ramp length= 0.8 ÷ sin 30 = 1.6 m
Work done = mg sin Θ * distance
Work done = 178 N * sin 30 * 1.6

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Work done = 142.4 Joules
c. Which situation requires more force?
Solution
Lifting the box vertically without using the ramp.
Explanation:
The sloping surface of the ramp supports part of the weight of the box as it moves
up the slope.
d. Which situation is distance moved greater?
Solution
Greater distance is moved when the ramp is used (1.6 m).
Calculations
Vertical distance when box is lifted perpendicularly= 0.8 m
Distance covered by the box when lifted through the ramp
Ramp length= 0.8 ÷ sin 30 = 1.6 m
Therefore, a greater distance is covered when using the ramp (1.6 m) than lifting
the box without using the ramp (0.8 m).
e. What is change in the gravitational potential of the box in each situation?
Solution:
The gravitational forces are equal
Calculations:
Lifting without a ramp
Work = force * distance

Module 4 12
Work = 178 N * 0.8 m
Work = 142.4 Joules
Lifting using a ramp
Ramp length= 0.8 ÷ sin 30 = 1.6 m
Work done = mg sin Θ * distance
Work done = 178 N * sin 30 * 1.6
Work done = 142.4 Joules
f. What advantage, if any, is there to using the ramp? Explain
Solution
The ramp makes work easier
Explanation:
The sloping surface of the ramp assists by supporting the weight of the
object unlike lifting directly as a result less force is required to move the load up
the ramp.