1) The document provides solutions to physics problems involving forces, velocities, accelerations, and displacements acting on objects with given masses. Force diagrams and calculations are shown. 2) One problem involves a barrel containing Batfink that is pushed along a cliff and its acceleration and final velocity before falling are calculated. 3) Another determines the tensions on chains suspending the Incredible Hulk, giving his mass and the angles and tension on one chain.

2. Know Definitions of Key
Terms & Symbols

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4. 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?

5. SOLUTION:
Force of gravity on Batfink and the barrel.
mbf = 50 kg
mb = 25 kg
g = -9.8 m/s2
Fg
Fg= -735 N

6. 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.

7. yy
Fs = 735 N
Ff = -2.75 N
x
Fa = 85 N
Fg = -735 N

8. Calculate the acceleration
of the barrel in the x
axis.

9. 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

10. What is the Vf of the
barrel just before it falls
off the cliff?

11. 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

12. Batfink and the barrel are raised at
1.25 m/s2. What is the force of
support acting on Batfink and the
Barrel?

13. 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

14. Suddenly, Batfink and the
barrel are lowered at .75
2
m/s . What is the force of
support acting on Batfink
and the Barrel?

15. 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

16. 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.

17. y
T1
T2 = 2500 N
26°
Fg -3479 N
32°

18. Determine the tension in T2X.

19. 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

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 = .34 m/s2
F = .935 N
Vf = 5.8 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:
may = 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