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. 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. 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. 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. yy
Fs = 735 N
Ff = -2.75 N
x
Fa = 85 N
Fg = -735 N
7. Calculate the acceleration
of the barrel in the x
axis.
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. What is the Vf of the
barrel just before it falls
off the cliff?
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. Batfink and the barrel are raised at
1.25 m/s2. What is the force of
support acting on Batfink and the
Barrel?
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. 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. SOLUTION:
Force of support on Batfink and the barrel.
mbf = 75 kg
mb = 25 kg
g = -9.8 m/s2
a = -.75 m/s2
Fs
Fs= 678.75
N
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. y
T1
T2 = 2500 N
26°
Fg -3479 N
32°
17. I dare you to determine the tension in
T2X.
18. SOLUTION:
Tension in T2x.
T2 = 2500 N
θ = 32°
m = 355 kg
g = -9.8 m/s2
Fg = -3479 N
T2x
T2x = 2120 N
19. Now determine
the tension in
T1!
20. 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
21. 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.
22. 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
23. 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.
24. 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 = .33 m/s2
F = .908 N
Vf = 2.99 m/s
25. 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?
26. SOLUTION:
4.14
6.2
max = Fa + Ff
(2.75)(.33) = Fa+ -0.75 N
Fa = 1.66 N
2.99
Fs = 26.95 N
1.47
Ff = -0.75 N
Fa = 1.66 N
.33
Fg = -26.95 N
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