The document discusses calculating accelerations and net forces for various objects including: - A 300 lb football player whose mass is calculated as 136 kg - An 86 kg man on the moon where gravitational force is 1/6 that of Earth - Forces required to accelerate vehicles like a 540 kg car and 2160 kg SUV from 0 to 60 mph in 10 seconds - Net force on a 158 kg piano being moved by two people against friction

1. Mrs. Reynolds

2. About twenty percent of the National Football
League weighs more than 300 pounds. At this
weight, their Body Mass Index (BMI) places
them at Grade 2 obesity, which is one step
below morbid obesity. Determine the mass of a
300 pound (1330 N) football player.

3. About twenty percent of the National Football
League weighs more than 300 pounds. At this
weight, their Body Mass Index (BMI) places
them at Grade 2 obesity, which is one step
below morbid obesity. Determine the mass of a
300 pound (1330 N) football player.
Fg = 300 lb. = 1330 N Fg = mg
g = 9.8 m/s^2 m = Fg/g
m=? m = 1330 N/9.8 m/s^2
m = 136 kg

4. According to the National Center for Health
Statistics, the average mass of an adult
American male is 86 kg. Determine the mass
and the weight of an 86-kg man on the moon
where the gravitational field is one-sixth that of
the Earth.
mearth = 86 kg
Gmoon = (1/6)g
= (1/6)*(9.8)
mmoon = ?
Fg moon = ?

5. According to the National Center for Health
Statistics, the average mass of an adult
American male is 86 kg. Determine the mass
and the weight of an 86-kg man on the moon
where the gravitational field is one-sixth that of
the Earth.
mearth = 86 kg mass doesn’t change
Gmoon = (1/6)g so mmoon = 86 kg
= (1/6)*(9.8)
mmoon = ?
Fg moon = ?

6. According to the National Center for Health
Statistics, the average mass of an adult
American male is 86 kg. Determine the mass
and the weight of an 86-kg man on the moon
where the gravitational field is one-sixth that of
the Earth.
mearth = 86 kg Fg moon = mmoon Gmoon
Gmoon = (1/6)g Fg moon = (86)(9.8/6)
= (1/6)*(9.8)Fg moon = 140 N
mmoon = ?
Fg moon = ?

7. a. Determine the net force required to
accelerate a 540-kg ultralight car from 0 to
27 m/s (60 mph) in 10.0 seconds.
b. Determine the net force required to
accelerate a 2160-kg Ford Expedition from 0
to 27 m/s (60 mph) in 10.0 seconds.

8. a. Determine the net force required to
accelerate a 540-kg ultralight car from 0 to
27 m/s (60 mph) in 10.0 seconds.
m = 540 kg
vi = 0
vf = 27 m/s
t = 10.0 s
Fnet = ?

9. a. Determine the net force required to
accelerate a 540-kg ultralight car from 0 to 27
m/s (60 mph) in 10.0 seconds.
m = 540 kg To find Fnet we need
vi = 0 mass and accel. We
vf = 27 m/s have mass, but not accel.
t = 10.0 s So we need to find accel.
Fnet = ?

10. 

11. a. Determine the net force required to
accelerate a 540-kg ultralight car from 0 to 27
m/s (60 mph) in 10.0 seconds.
m = 540 kg Fnet = ma
vi = 0 Fnet = (540 kg)(2.7 m/s^2)
vf = 27 m/s Fnet = 1458 N
t = 10.0 s
a = 2.7 m/s^2
Fnet = ?

12. b. Determine the net force required to
accelerate a 2160-kg Ford Expedition from 0
to 27 m/s (60 mph) in 10.0 seconds.
m = 2160 kg
vi = 0
vf = 27 m/s
t = 10.0 s
Fnet = ?

13. 

14. b. Determine the net force required to
accelerate a 2160-kg Ford Expedition from 0 to
27 m/s (60 mph) in 10.0 seconds.
m = 2160 kg Fnet = ma
vi = 0 Fnet = (2160 kg)(2.7 m/s^2)
vf = 27 m/s Fnet = 5832 N
t = 10.0 s
a = 2.7 m/s^2
Fnet = ?

15. The Top Thrill Dragster stratacoaster at Cedar
Point Amusement Park in Ohio uses a hydraulic
launching system to accelerate riders from 0 to 54
m/s (120 mi/hr) in 3.8 seconds before climbing a
completely vertical 420-foot hill . Determine the net
force required to accelerate an 86-kg man.
vi = 0 m/s Fnet = ?
vf = 54 m/s
t = 3.8 s To find Fnet we need to find
d = 420 ft acceleration.
m = 86 kg

16. 

17. The Top Thrill Dragster stratacoaster at Cedar
Point Amusement Park in Ohio uses a hydraulic
launching system to accelerate riders from 0 to 54
m/s (120 mi/hr) in 3.8 seconds before climbing a
completely vertical 420-foot hill . Determine the net
force required to accelerate an 86-kg man.
vi = 0 m/s Fnet = ?
vf = 54 m/s
t = 3.8 s
d = 420 ft
m = 86 kg
a = 14.2 m/s^2

18.  Blake is dragging a bag of grass from the
garage to the street on the evening before
garbage pick-up day with a force of 40.2 N.
The frictional force between the bag and
the ground is 5.7 N. What is the net force
acting on the bag?

19.  Blake is dragging a bag of grass from the
garage to the street on the evening before
garbage pick-up day with a force of 40.2 N.
The frictional force between the bag and
the ground is 5.7 N. What is the net force
acting on the bag?
5.7 N 40.2
N

20. Silas and Ivey are members of the stage crew
for the Variety Show. Between acts, they must
quickly move a Baby Grand Piano onto stage.
After the curtain closes, they exert a sudden
forward force of 524 N to budge the piano from
rest and get it up to speed. The 158-kg piano
experiences 418 N of friction.
a. What is the piano's acceleration during this phase of
its motion?
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?

21. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
a. What is the piano's acceleration during this phase of
its motion?
F1 = 524 N forward
F2 = 418 N backward
m = 158 kg
a=?
418 N 524 N

22. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
a. What is the piano's acceleration during this phase of
its motion?
F1 = 524 N forward Fnet = 524 N – 418 N = 106 N
F2 = 418 N backward
m = 158 kg
a=?
418 N 524 N

23. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
a. What is the piano's acceleration during this phase of
its motion?
F1 = 524 N forward Fnet = 524 N – 418 N = 106 N
F2 = 418 N backward Fnet = ma, so a = Fnet/m
m = 158 kg a = (106 N)/(158 kg)
a=? a = 0.671 m/s^2

24. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?

25. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?
a = 0.671 m/s^2
t = 1.44 s

26. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?
a = 0.671 m/s^2
t = 1.44 s
vi = 0 m/s
vf = ?

27. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?
a = 0.671 m/s^2
t = 1.44 s
vi = 0 m/s
vf = ?

28. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?
a = 0.671 m/s^2 vf = vi + at
t = 1.44 s
vi = 0 m/s
vf = ?

29. They exert a sudden forward force of 524 N to
budge the piano from rest and get it up to
speed. The 158-kg piano experiences 418 N of
friction.
b. If Silas and Ivey maintain this forward force for 1.44
seconds, then what speed will the piano have?
a = 0.671 m/s^2 vf = vi + at
t = 1.44 s vf = 0 + (0.671 m/s^2)(1.44 s)
vi = 0 m/s vf = 0.966 m/s
vf = ?