Monday 18 Oct

Target: Understanding Motion

Test Today & Tomorrow (18 & 19 Oct)

Correct Free Responces:

Review for test

Pre Test  Ch 4 – 6 (Answers)                              

1.  What is the magnitude of the force needed to accelerate a 250 kg cart from rest to 5 m/s in 2.0 s?

F_net = ma  à  (250)(5) =  625N

2.       If the weight of an object is 20N and floor of an elevator causes a vertical acceleration is 20.0 m/s², what is the object’s apparent weight (normal force on the object)?

F_net = ma  à  (2)(20) = 40, so apparent weight is  60N

3.       Two boxes (A & B) are connected by a lightweight cord and are resting on a frictionless table.  The boxes have mass 20 kg and 5 kg, respectively with a line in-between the boxes.  A horizontal force of 100 N is applied to the 20 kg box.  a.) What is the acceleration of box B (5 kg)?  b.) Tension in line in-between the boxes

a. F_net = ma  à  100N = (25kg)(a)   à a = 4

b. F_net = ma  à  T – 0  = (5kg)(4)   à  T = 20N

4.       A 5 kg object is pushed up a 50° incline at a constant speed applying a force parallel to the slope.  The force of  kinetic friction is:

F_X = F_{k fric} = (Sinq)(Weight)  à  F_{k fric} = (.766)(50N)   à F_{k fric}   = 38.3N

5.       A frictionless, weightless pulley is used to connect the motion of two masses as shown.  The static coefficient of friction of m₁ and the table is 0.24.  If M₁ = 20 kg, how large must m₂ be to cause to start moving?

M₁

                                                                              m = 0.24

       F_{k fric} = (m)(F_N)  = (.24)(200) = 48N                                                                                          M₂                                                

6.       Three masses of 20 kg, 40 kg and 20 kg respectively are connected as shown below. They are pulled by a force (T₁) of 240N on a frictionless table. What is the tension T₂?

                                                                     T₁            20kg           T₂        40kg           T₃      20kg   

F_net = ma  à  240N = (80kg)(a)   à a = 3

F_net = ma  à  T₁ – T₂  = (20kg)(3)   à   T₂  = 240N - (60N)  à   T₂ = 180N

7.       A block of mass 60M can move with coefficient of friction (0.5) on a horizontal table.  This block is attached to another block of mass 40M by a cord that passes over a frictionless pulley, as shown.  a.) What is the magnitude of the acceleration of the descending block? B.) Find tension.

                            60M        

        a. F_net = ma  à   W_40M - F_{k fric} = ma

                   400N - (m)(F_N)  = (100kg)(a)

                    400N - (0.5)(600)  = (100kg)(a)                                                  40M

                           a = 100N/100kg à a = 1.0m/s²

b. F_net = ma à  W – T  = (40kg)(1.0) à T = 400 – 40 à  T = 360N                              

8.       The two ends (A & B) of a 20.0 kg plank rest on scales.  A 60.0 kg mass is placed 1/3 of the way in from A end.  The two scales register weights of:

CCW Torque =  CW Torque

   (F_B)(X_B)      = (F_beam)(X_beam) + (F_mass)(X_mass)

   (F_B)(X)    = (200N)(0.5X) + (600N)(0.333X)

    (F_B)        =        (100N)       +   (200N)

        F_B    =      300N

           F_A     =  800N – 300N = 500N

9.        If spacecraft A  has three times the mass of spacecraft B  (both in the same location)

a. Compare mass, Weight (a & b) when in deep space.

b. Compare centripetal acceleration, centripetal force  & Velocity if both in same orbit around the Earth

a.  Mass of  A     =  3B mass,   No weight for either (i.e., both = 0)

b.             F_c = m_Aa_c     à        3F_c = 3m_Ba_c                      

a_c: for both object in orbit are = (mass isn’t important, all objects fall at the same rate).

F_c: A has 3 times the mass, needs 3 times the force to provide the same a_c 

V: for both object in orbit are = (see a_c: above)

10.    A 500 kg mass is tied to a line which has a breaking strength of 20 kN. a.) What is the centripetal acceleration (in a horizontal circle with a 2.0 m radius)? b. What is maximum speed the mass can have before breaking the line?

a.  F_c = m_Aa_c à 20,000N = (500)( a_c)  à a_c  = 40 m/s²

b.  a_c = V²/r à 40 m/s² = (V²)/(2m)  à v = sq rt of 80 = 8.9 m/s

11.     Whirling a ball tied to the end of a string in a horizontal circle and quadrupling its speed a.) What happens to its centripetal acceleration b.) centripetal force?

_a.        a_c = V²/r à quadrupling its speed you 16 times its a_c

b.       F_c = m_Aa_c à 16 times its a_c,  its F_c is also increased by 16 times

12.    Planet X has a radius of 1/3 and a mass 3 times that of the Earth.  The acceleration of gravity on the Planet x is (in m/s):

Gravity on Earth =  m_E / r_E²   

Gravity on X =  3m_E / (1/3 r_E² )

à Planet X has 27 times the gravity of Earth, so (27 x 10 =) 270 m/s²

13.    A 40N box is pushed up a 20° inclined ramp at a constant speed by a horizontal force.  The work done after moving the box 100 m up the inclined ramp is:

  W (input)  =   PE + KE  (output)

  W = force x dist   or  if against gravity = mgh (if no acceleration)

  W = mgh = (4)(10)(sin 20)(100)  = 1368 J

14.    If (with no friction) a force a force F results in an acceleration a when acting on a mass m. Then a mass of 5m and a force 10F will result in an acceleration of

F_net = ma  à  10F_net = 5ma₂ à  a₂ = 2a

15.    A trunk is pulled across a horizontal floor at a constant speed by a 100N force applied by a rope at an angle of 30° above the horizontal.  The work done by friction after moving the trunk 5.0 m is:

W = force x dist   à (cos30⁰)(100N)(5) = 433J

16.    Two boxes slide down frictionless incline planes from a height of 20m.  First at a 30⁰ angle and the other at a 70^0.

a.) Speeds of each at the bottom?  

equal because both lost equal PE à mgh = ½ mv²

à (10)(20m)  = ½ v²

à (sq root of 400)  = V = 20m/s

b.) Fx and Fy for each.

30^0:       Fx = (sin30⁰)(mg)  =   0.5(mg)

                Fy = (cos30⁰)(mg)  =  0.87(mg)

70^0:       Fx = (sin70⁰)(mg)  =  0.94(mg)

                Fy = (cos70⁰)(mg)  =  0.34(mg)

c.) Accelerations for each (4kg)

30^0:       Fx = (sin30⁰)(mg) à  20N = ma à a = 5 m/s²

70^0:       Fx = (sin70⁰)(mg) à  37.6N = ma à a = 9.40 m/s²

d.) Time to reach bottom for each

Dist down the plane:

 30^0:      Hyp = Height / sin30⁰  = 40m

70^0:       Hyp = Height / sin70⁰  = 21.3m

                30^0:      x =  (V_o t) + ½ (at ²)  à  40m = 0 + ½ (5)(t²)   à t = 4.0 sec

           70^0:      x =  (V_o t) + ½ (at ²)  à  23.1m = 0 + ½ (9.4)(t²)   à t = 2.22 sec

17.    How much work is required to accelerate a 50 kg car from 0 m/s to 20 m/s?

W = KE + PE   à   ½ mv²  + 0   à  ½ (50)(20²)  à W = 10,000 J

18.    If they continue to accelerate the car (in the above question #17) from 20 m/s to 40 m/s, Find:

      a.) How much work?

W = KE + PE   à  (½ mv² )_f – (½ mv² à  (½ mv² )_i

à W = 40,000 -10,000 = 30, 000 J

      b.)  If you double the velocity, does it take twice the work? Explain

No. V is squared so it takes 4 times the amount of work.

19.    An object from a planet that has no atmosphere.  The object falls freely for 50 meters during the first 5 seconds.  What is the magnitude of the acceleration due to gravity on the planet?

x =  (V_o t) + ½ (at ²)  à 50m = 0 + ½ (a)(5²)

      a = 4 m/s²

20.    How fast must a 4.0 kg mass be moving if its kinetic energy is 200 J?

    KE =   ½ mv²    à   200J = ½ (4)v^{2   à} V = sq root of 100 = 10 m/s

21.    A large locomotive exerts an average force of 3,000N while pulling a train from rest a distance of 2 km.  How much work does it do on the train during this event?

                                W = force x dist   à  W = (3,000)(2,000)  = 6 x 10⁶J

22.    Gun is fired at a 45⁰ angle. Described the series of energy transformations as the bullet travels from the gun to the ground.

Chem eng (gun power) à Mech eng (bullet moving in to out of gun) à  PE (partial as it gains height) à back into all KE as it hits the ground

23.    A wagon rolls from rest down a hill reaching the bottom at 20 m/s.  On the next run,

the wagon gets a push and starts down at 30 m/s.  At what speed does it now have at the bottom?

case 1:                    PE₁ + KE₁  = PE₂ + KE₂

                        mgh₁ + 0  = 0 + ½ mv²

      10h₁ + 0  = 0 + ½(20)² à  h = ½ (400)/10  = 20m

case 2:                    PE₁ + KE₁  = PE₂ + KE₂

                        mgh₁ + ½ mv₁²  = 0 + ½ mV₂²

               (10)(20) + ½ (30²)  = 0 + ½ V₂²

      à sq root { (650)(2) }  = V₂   =   36 m/s

24. A rocket coasting along in space at some speed V, fires its engines thereupon ½ of its speed, but at the same time it jettisons some cargo, reducing its mass to half its previous value.  a.) What is new momentum compared to the old.   b). In the process, its KE is (compared to the original):

(M₁)(V₁) à (0.5 M₁)(0.5V₁)  = 0.25 original

25.     A ball as it rolls down the circularly curved track as shown, its speed, acceleration and kinetic energy (KE), respectively
                                                    

a.       Speed  increases

b.       Acceleration  decreases

c.        KE increases

A ball of mass m is dropped from rest at a height H above the ground from rest and bounces back in the positive upward direction to its height H after a perfectly elastic collision with the metal plate on the ground.

26.    Draw a velocity, Potential Energy(PE), and Kinetic Energy(KE) vs. time graphs as a function of:

      a. height (y)

        b. time (t)

         V                          PE                          KE

                                    y                     y                        y

         V                            PE                      KE

                                    t                         t                 t

27.    A space shuttle of mass M circles around the earth with constant speed V. 

Does its weight disappears in its orbit because the centripetal force and the gravitational force on it cancel each other? (Explain)

                No/neither;  its in free fall

28.    The mass of a car is 200 kg.  The constant power supplied by the engine is 50 kW and the car starts from rest.    At the end of 30 seconds, Find

a.       Work  =  (force)(dist) or (Power)(time)

        (50,000)(30)  = 1,500,000 J

b.       Final velocity:

KE = ½ mv₁²  à  1,500,000 = ½(200)( v²)

                                à  sq root 15,000 = v = 122.5 m/s

c.        KE: Same as Work  1,500,000 J

d.       Change in momentum (Dr)

(m)(Dv)  = (200)(122.5)  = 24,500 kgm/s

e.       Impulse

(F)(t)  = (m)(Dv)  à = 24,500 kgm/s

f.         Force Applied

                                (F)(t)  = (m)(Dv) à   (F)(30)  = (200)(122.5) à F = 817N

29.    An  800kg car is traveling in a horizontal direction with a speed of 20 m/s at time t = 0 s.
The force F of 500N is applied till the car is stopped. Find:

                a. Momentum (initial) (m)(v)  =  (800)(20) = 16,000 kgm/s

                b. Impulse (F)(t)  = (m)(Dv) à  (800)(20) = 16,000 kgm/s

                c. Time to stop (F)(t)  = (m)(Dv) à (500N)(t) = (800)(20) à 32s       

                d. Acceleration   a = (Dv)/t  = 20/32  =   0.625 m/s²

              e. Work to stop KE = ½ mv₁²  à  ½(800)(20²) à 160,000 J

                f. Distance Traveled to stop

x =  (V_o t) + ½ (at ²)  à  x = (20)(32) - ½ (0.625)(32²)  

                                                642 -320   = 320m

                g. Stopping Power:    power = w/t  à 160,000/32 = 5,000W

31. A ball (5.0 kg) drops from a height of 5.0 m and bounces back up 4.0 m.  The change in momentum (Dr) of the ball is: 

Down 5m: V₂² – V₁² = 2aDx  à V₂² - 0 = 2(-10)(5)  à V = Sq root of 100 = -10m/s

Down 4m: V₂² – V₁² = 2aDx  à V₂² - 0 = 2(-10)(4)  à V = Sq root of 80   = +8.94 m/s

DMo (or Dr) = mDV  à (5kg)(18.94m/s)  = 94.7 (kgm/sec

30.    A ball falls straight down through the air under the influence of gravity.  There is a retarding force F on it with a magnitude given by F = bv, where v is the speed and b is a positive constant.  The magnitude of the acceleration is equal to which of the following?

1.                                                       g – b

2.                                                       g – bv/m

3.                                                       g + bv/m

4.                                                       g/b

5.                                                       bv/m

                        F_net  =  ma 

      mg – bv =  ma

                a  = (mg/m) – (bv/m)   =  g - (bv/m)  

31.    A mass m on a string of length R is released from rest from the horizontal position.   At its lowest position the tension in the string would be (show work):

1.                                                      mg downward

2.                                                      3 mg downward

3.                                                      2 mg upward                                                                    R    

4.                                                      3 mg upward

5.                                                      mg/2 upward                                      F_c=  T –  mg

#4

mgh = ½ mv²  à v² = {2gh}  or {2gR}                                       F_c= m(v²)/(R)

 F_c  =  T – mg = m(v²)/(R)

          T = m{v²}/(R) + mg

      à =m{2gR}/(R)  + mg

            T =  2mg + mg = 3mg