10 2012 ppt force review

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 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.

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 = -.75 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

I dare you to 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

Now 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 = 1.66 N

Fa = 1.66 N

6.2

4.14

1.47

2.99

.33