AP Physics: Thurs SEP 24:

Target:

Learning Objective: Impulse & Mechanical Advantage

Journal: 

 

    A lump of soft clay of mass m falls from a height h above the floor, where it crashes and comes to rest. 1. What force brings the clay to a stop? 2. How much work is done by the floor on the clay? 3. Where does that energy end up?

 

 

 

 

 

 

 

 

 

 

 

1. Force from the floor

2. W = Dmgh

3.  Sound, D temp of clay and floor

2.

 

                 

                        5000m

                              100

 

A 100,000 kg train climbs a 5,000 m mountain track at a speed of 20m/s at an angle of 10˚ with respect to the horizontal.

1. How much altitude does it gain?

            a. 868 m      b. 1230 m

            c. 3410 m    d. 4920 m

2. How much work was done?

 

3. How much power was used?

 

4. What is the mechanical advantage (MA) of the mountain?

 

 

 

 

 

 

1. Sin10˚(5000m)  = 868 m

2. W = (f)(x) = DPE

     W = Dmgh 

  = (100,000kg)(10N/kg)(868m)

   = 8.68 x 108 J

3. P = W/t 

  P = 8.68 x 108 J / (5000m/20m/s)   P = 3.47 x 106 W

4. MA = ratio of distances

           =  5000m/868m = 5.76

 

                

Chapter 4:

(forces)

 

  

*Free-body diagrams & Forces:

 

 

1. The three blocks are accelerating at 5m/s2 Draw Free Body diagrams & find the tensions in the lines.

 

 

 

 

 

 

 

 

 

 

 

 

T1 , T2 , & T3

 

Fnet = ma   à F = (60kg)(5N/kg)à 300N

  so,  T1= 300N

 

 

 Fnet on 30kg block

               =  T1 - T2  = ma  

  à Fnet = 300N - T2  = (30kg)(5N/kg)

  à 300N - 150N = T2

  so,  T2= 150N

 

 

Fnet on 10kg block

              =  T2 - T3  = ma  

  à Fnet = 150N - T3  = (10kg)(5N/kg)

  à 150N - 50N = T2

  so,  T2= 100N

 

 

 

2. What is the

a. Mechanical Advantage 

            b. Effort Force (Resistance force is 100N)

c. Effort Distance (Resistance dist. is 2.0m)

                        I.          incline plane of 300

                        II.        Wheel (2.4m) & Axel (0.2m)

                        III.       Pulley of 1 weight bearing line

                        III.   Pulley of 3 weight bearing lines

                        IV.       Pulley of 4 weight bearing lines

V.             12m Lever: 2m on resistance side of   fulcrum and 10m on effort side

VI.         16m incline plane that is 4m high

 

I. a.  Sin 300 = .5,    ratio of distances

       so MA = 2

            b.  50N  (½ the force)

            c. 4.0 m (twice the resistance distance)

 

II.                a.  2.4/0.2 = 12, so MA = 12

b. 8.33N     (100/12)

c. 24m       (2m)(12)

 

III.             a. MA = 1

b.      100N

c.       2m

 

IV.              a. MA = 4

b. 25N       (100/4)

c. 8m         (2m)(4)

 

V.                 a. MA = 5

b.      20N     (100/5)

c.       10m     (2m)(5)

 

VI.              a. MA = 4

b. 25N       (100/4)

c. 8m         (2m)(4)

 

 

 

 

3. A 170kg hot air balloon hovers at a constant altitude above the ground. The gondola while a 80kg person is hanging from a rope below the gondola. The person on the rope starts to ascend at constant rate at 1.0m/s.

a. What is the direction of the gondola as he ascends?

b. Speed of the balloon as he ascends? 

b. If the person climbs for 10 seconds and then stops, what is the change in the balloons altitude?

 

a & b.)

Momentum (Mo) is conserved

     Mo up = Mo down 

mVup =  mVdown à (80)(1) = (170)(Vdown)

Vdown = - 0.471 m/s

b.  Person below went up 10m: 

PE is conserved (no KE)

mghperson =    mghgondola

(80)(10)(10) = (170)(10)(y)

y = 4.71m down

 

2nd way also:

Distance = Vt = (0.471m/s)(10s) =  4.71m down

 

 

 

4. A trunk is pulled across a floor at a constant speed by a 200N force applied by a rope at an angle of 600 above the horizontal. If the trunk weighs 150N. The coefficient of friction is:

 

 

 

 

 

 

 

 

 

 

 

 

FK = Fhorz  = (m) FN

(m) FN     =      (Cos 60)(200)

(m) (mg - (Sin 60)mg = (Cos 60)(200)

 (m) (150 - .87(150) = (.5)(200)

(m) (150 - 130) = (.5)(200)

  m = 100 / 20  = 5

 

 

5. A man pushes a 10kg cart up a 300 incline at a constant speed. He applies  60N force parallel to the slope.

a. What is the Normal force?

b. What is the force down the plane?  

c. What is the force of Kinetic friction?

d. What is kinetic m?

                                                                                        

FN = (Cos q)(mg)

 

 Fx = (Sin q)(mg)

 

 

 

 

a. FN = (Cos 30)(mg)  = (.87)(10)(10) = 87N

b. Fx = (Sin 30)(mg) = (.5)(10)(10) = 50N (no friction)

c. Constant speed means no acceleration (Balanced forces)

    Applied force = Fx  + Friction

        60N =  50N + 10N (of Kinetic friction)

d. FK = mFN = m (Cos 30)(mg)

FK = 10N = m (87N)

m = .115

 

 

 

6. The block is resting on an incline plane (static m = 0.30 and kinetic m = 0.18). The angle is increased until the block starts to move.

  a. What is the angle (block starts to move)?

  b. What is the block’s acceleration at this angle?

c. What is the speed of the block when it hits the  ground (the length of the incline plane = 4.0m)?

d. After reaching the floor, the block slides with    the same kinetic m (0.18). How far does it slide?

 

Fx = (Sin q)(mg):    FS = mSFN:     FN =  (Cos q)(mg)

    a)               Fx = FS  

à    (Sin q)(mg) =  mS(Cos q)(mg)

à    (Sin q)   =  mS(Cos q) 

à     mS = (Sin q)/(Cos q) =  Tanq

  mS = Tanq

        .3 = Tanq

à  Tanq -1 (.3) = 16.70

 

b. Fx = (Sin q)(mg) ,    FK = mKFN ,     FN =  (Cos q)(mg)

     Fnet     =   ma          

  Fx – FK = ma  

(Sin q)(mg) –  mK(Cos q)(mg)  =  ma

à  (Sin 16.7)(g)  –  0.18(Cos 16.7)(g)  =  a

à     (.287)(10)  –  0.18(.958)(10)  =  a

à     (2.87)  –  (1.72)  =  a  à a = 1.15m/s2

 

 

 

 

 

  1. Vf2 – Vo2 = 2aDs

Vf2 – 0 = 2(1.15)(4)

Vf = √ 9.2  = 3.03m/s

  

  1. Fnet = ma          à       FK = ma

mK(mg) = ma  à  mK(g) = a 

a = 0.18(10) =     -1.8m/s2   

 Vf2 – Vo2 = 2aDs

 0 – (3.03)2 = 2 (-1.8)s   à s = 5.6m

 

 

            a                                 T =              

                                                      4kg                           

 

  


 

                                                                                       5kg

 

 

12. A 5kg block hangs over a table and pulls a 4kg block on the table with  mK = 0.4. What is the:

a. Net force?

            b. acceleration?

            c. tension in line?

 

  1. Fnet =   W5 - FK

      Fnet  =  m5g – (mm4g)

            Fnet = 50  -  (.4)(40) à 16N

            Fnet = 50  -  16N = 34N

    

 

  1. Fnet =   mtotala

      34   =   9a   à a = 3.8m/s2

 

  1.  Fnet =   m5a

      W – T = m5a

      50 –T = (5)(3.8)

      T = 50 - 18.9  à 31.1N

 

13. A 10.0m board (100kg) is hanging 2.5m over the edge of a platform (that is above the water). Jimmy wants to take a good look at the water. If Jimmy (120kg) starts walking towards the end of the board, will he make it all the way without falling in? If not, how far can he walk on the board before both he and the board fall into the water below? 

          Jimmy M.            

  


 

Cloud Callout: I’m frightened

                                            

 

  

 


 

                

 

  


 

                           7.5m             2.5m

                                                                                          

GOC of board is 5m à 2.5m from edge

Torque from platform edge: 

fd (ccw) = fd (cw)

(1000)(2.5) = 1200(x)

x = 2.08m                                             

Yes, sorry to say, Jimmy would fall!