Know Definitions of Key
Terms & Symbols
Focus during the entire Power Point activity.
Solidify your studying skills during this
class period.
Perform your work in...
Batfink, who has a mass of 50 kg is
placed in a 25 kg stationary barrel.
What is the Fg on Batfink and the
barrel?
SOLUTION:
Force of gravity on Batfink and the barrel.

mbf = 50 kg
mb = 25 kg
g = -9.8 m/s2

Fg

Fg= -735 N
Hugo Ago-go pushes the barrel with
Batfink in it towards the end of the
cliff with a 85 N force over a distance
of 12 m be...
yy
Fs = 735 N
Ff = -2.75 N

x
Fa = 85 N
Fg = -735 N
Calculate the acceleration
of the barrel in the x
axis.
SOLUTION:
Acceleration of Batfink and the barrel.

mbf = 5o kg
mb = 25 kg
g = -9.8 m/s2
Fa = 85 N
Ff = 2.75 N
xi = 0 m
x =...
What is the Vf of the
barrel just before it falls
off the cliff?
SOLUTION:
Final velocity of Batfink and the barrel while
still on the cliff.
vi = 0 m/s
a = 1.1 m/s2
xi = 0 m
xf = 12 m
ti...
Batfink and the barrel are raised at
1.25 m/s2. What is the force of
support acting on Batfink and the
Barrel?
SOLUTION:
Force of support on Batfink and the barrel.

mbf = 75 kg
mb = 25 kg
g = -9.8 m/s2
ay = 1.25 m/s2

Fs

Fs= 828.75...
Suddenly, Batfink and the
barrel are lowered at .75
2
m/s . What is the force of
support acting on Batfink
and the Barrel?
SOLUTION:
Force of support on Batfink and the barrel.

mbf = 75 kg
mb = 25 kg
g = -9.8 m/s2
a = -1.1 m/s2

Fs

Fs= 678.75
...
The Incredible Hulk is hanging motionless off
the ground by chains attached separately to
his wrists from two different wa...
y
T1
T2 = 2500 N
26°

Fg -3479 N

32°
Determine the tension in T2X.
SOLUTION:
Tension in chain #1. First find T2x.

T2 = 2500 N
θ = 32°
m = 355 kg
g = -9.8 m/s2
Fg = -3479 N

T2x

T2x = 2120...
SOLUTION:
Tension in chain #1.

T2 = 2500 N
θ = 26°
m = 355 kg
g = -9.8 m/s2
Fg = -3479 N
T2x = 2120 N

T1

T1 = 2358.2 N
Determine the force, velocity and
displacement for a 2.75 kg cart starting
6.2 meters left of the reference point,
while t...
SOLUTION:
Rate of acceleration, force, velocity, and
final displacement.
a = 2 m/s2
F=
m = 2.75 kg Vf =
t = 4.63 s
Xf =
Vi...
Determine the force, velocity and
displacement for a 2.75 kg cart starting
6.2 meters left of the reference point,
while t...
SOLUTION:

Rate of acceleration, force, velocity, and
final displacement.
Xi = -6.2 m
F=
Vi = 1.47 m/s a =
m = 2.75 kg Vf ...
Create a motion map, properly labeled x-t,
v-t, and a-t graphs, and a force diagram
based on the actual F-m graph assuming...
SOLUTION:
max = Fa + Ff
(2.75)(.34) = Fa+ -0.75 N
Fa = 1.69 N
Fs = 26.95 N

Ff = -0.75 N

Fa = 1.69 N
Fg = -26.95 N
10 2012 ppt force review
10 2012 ppt force review
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10 2012 ppt force review

  1. 1. Know Definitions of Key Terms & Symbols
  2. 2. Focus during the entire Power Point activity. Solidify your studying skills during this class period. Perform your work in your science journal so you have created a study guide for the test. Call me over if you are having difficulty getting started. If your answer is confirmed as correct, become a student/teacher and help someone in class who does not understand the method used to solve the problem.
  3. 3. Batfink, who has a mass of 50 kg is placed in a 25 kg stationary barrel. What is the Fg on Batfink and the barrel?
  4. 4. SOLUTION: Force of gravity on Batfink and the barrel. mbf = 50 kg mb = 25 kg g = -9.8 m/s2 Fg Fg= -735 N
  5. 5. Hugo Ago-go pushes the barrel with Batfink in it towards the end of the cliff with a 85 N force over a distance of 12 m before the barrel leaves the cliff. The force of friction is 2.75 N. Draw a force diagram of the situation.
  6. 6. yy Fs = 735 N Ff = -2.75 N x Fa = 85 N Fg = -735 N
  7. 7. Calculate the acceleration of the barrel in the x axis.
  8. 8. SOLUTION: Acceleration of Batfink and the barrel. mbf = 5o kg mb = 25 kg g = -9.8 m/s2 Fa = 85 N Ff = 2.75 N xi = 0 m x = 12 m a a = 1.1 m/s2
  9. 9. What is the Vf of the barrel just before it falls off the cliff?
  10. 10. SOLUTION: Final velocity of Batfink and the barrel while still on the cliff. vi = 0 m/s a = 1.1 m/s2 xi = 0 m xf = 12 m ti = 0 s vf tf v = 5.14 m/s
  11. 11. Batfink and the barrel are raised at 1.25 m/s2. What is the force of support acting on Batfink and the Barrel?
  12. 12. SOLUTION: Force of support on Batfink and the barrel. mbf = 75 kg mb = 25 kg g = -9.8 m/s2 ay = 1.25 m/s2 Fs Fs= 828.75 N
  13. 13. Suddenly, Batfink and the barrel are lowered at .75 2 m/s . What is the force of support acting on Batfink and the Barrel?
  14. 14. SOLUTION: Force of support on Batfink and the barrel. mbf = 75 kg mb = 25 kg g = -9.8 m/s2 a = -1.1 m/s2 Fs Fs= 678.75 N
  15. 15. The Incredible Hulk is hanging motionless off the ground by chains attached separately to his wrists from two different walls. The Hulk has a mass of 355 kg. The chain on his right wrist (T1) forms an angle of 26˚ relative to the floor, and the chain from his left wrist (T 2) forms an angle of 32˚ relative to the floor. T2 has 2500 N acting on it. Draw a force diagram of the situation.
  16. 16. y T1 T2 = 2500 N 26° Fg -3479 N 32°
  17. 17. Determine the tension in T2X.
  18. 18. SOLUTION: Tension in chain #1. First find T2x. T2 = 2500 N θ = 32° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x T2x = 2120 N
  19. 19. SOLUTION: Tension in chain #1. T2 = 2500 N θ = 26° m = 355 kg g = -9.8 m/s2 Fg = -3479 N T2x = 2120 N T1 T1 = 2358.2 N
  20. 20. Determine the force, velocity and displacement for a 2.75 kg cart starting 6.2 meters left of the reference point, while traveling at 1.47 m/s for 4.63 seconds.
  21. 21. SOLUTION: Rate of acceleration, force, velocity, and final displacement. a = 2 m/s2 F= m = 2.75 kg Vf = t = 4.63 s Xf = Vi = 1.47 m/s Xi = -6.2 m F = 5.5 N Vf = 10.73 m/s X = 22.04 m
  22. 22. Determine the force, velocity and displacement for a 2.75 kg cart starting 6.2 meters left of the reference point, while traveling at 1.47 m/s for 4.63 seconds.
  23. 23. SOLUTION: Rate of acceleration, force, velocity, and final displacement. Xi = -6.2 m F= Vi = 1.47 m/s a = m = 2.75 kg Vf = t = 4.63 s Xf = a = .34 m/s2 F = .935 N Vf = 3.04 m/s
  24. 24. Create a motion map, properly labeled x-t, v-t, and a-t graphs, and a force diagram based on the actual F-m graph assuming a force of friction of 0.75 N. What would the applied force have to be to attain this rate of acceleration?
  25. 25. SOLUTION: max = Fa + Ff (2.75)(.34) = Fa+ -0.75 N Fa = 1.69 N Fs = 26.95 N Ff = -0.75 N Fa = 1.69 N Fg = -26.95 N
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