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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.
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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.
Batfink, who has a mass of 50 kg isplaced 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 Fg
mb = 25 kgg = -9.8 m/s2
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 before the barrel leaves the cliff. The force of friction is 2.75 N. Draw a force diagram of the situation.
y y
Fs =
735 N
Ff = -2.75 N x
Fa = 85 N
Fg = -735 N
Calculate the acceleration
of the barrel in the xaxis.
SOLUTION:Acceleration of Batfink and the barrel.
mbf = 5o kg amb = 25 kgg = -9.8 m/s2
Fa = 85 NFf = 2.75 Nxi = 0 mxf = 12 m a = 1.1 m/s2
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 vf
a = 1.1 m/s2 tf
xi = 0 mxf = 12 mti = 0 s v = 5.14 m/s
Batfink and the barrel are raised at1.25 m/s2. What is the force ofsupport acting on Batfink and theBarrel?
SOLUTION:Force of support on Batfink and the barrel.
mbf = 75 kg Fs
mb = 25 kgg = -9.8 m/s2
ay = 1.25 m/s2
Fs= 828.75 N
Suddenly, Batfink and thebarrel are lowered at .75m/s2. What is the force ofsupport acting on Batfinkand the Barrel?
SOLUTION:Force of support on Batfink and the barrel.
mbf = 75 kg Fs
mb = 25 kgg = -9.8 m/s2
a = -1.1 m/s2
Fs= 678.75 N
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 (T2) forms an angle of 32˚ relative to the floor. T2 has 2500 N acting on it. Draw a force diagram of the situation.
y
T1 T2 = 2500 N
26° 32°
Fg -3479 N
Determine the tension in T2X.
SOLUTION:Tension in T2x.
T2 = 2500 N T2x
θ = 32°m = 355 kgg = -9.8 m/s2 Fg = -3479 N T2x = 2120 N
Determine the tension in T1.
SOLUTION:Tension in chain #1.
T2 = 2500 N T1
θ = 26°m = 355 kgg = -9.8 m/s2 Fg = -3479 NT2x = 2120 N 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 traveling at 1.47 m/s for 4.63
seconds.
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/sXi = -6.2 m F = 5.5 N
Vf = 10.73 m/s Xf = 22.04 m
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.
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 Xf = 4.14 m
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?
SOLUTION:
Fg = -26.95 N
Fs = 26.95 N
Ff = -0.75 N
max = Fa + Ff
(2.75)(.33) = Fa+ -0.75 NFa = 1.69 N
Fa = 1.66 N