Target: Understanding Motion

 

Equation Test on Friday 16th

Quiz (Equations 1-18a, 41-44) :    

Name _________________________________

 

1.      (Velocity after a period of time)

V_f=

 

2.      (Distance after a period of time)

X_f =

 

3.     (Velocity/distance/acceleration without time)

V_f²            =

 

4.       (center seeking acceleration or centripetal acceleration)

A_c=

 

5.     Work in relationship with KE + PE (usingfundamental terms)

W  =

 

6.      (Change of momentum (Dr )=  Impulse)

 

            =

 

7.      (Power with work)

P_avg =

 

8.      (Power with veolcity)

P =

9.      (Force exerted by a springy)

F_s =

 

10.  (Potential energy of a spring)

U_s =

 

11.  (period of a spring)

T_s =

12.  (period of a pendulum)

T_p =

13.  (relationship between period and frequency)

T =

14.  (Newton’s Law of Gravitation)

F_g =

 

 

15.  Pressure (with atmosphere) of a liquid

    P =

 

16.  (Force of buoyancy)

F_buoy =

 

17.  (Equation of continuity for a liquid)

A₁V₁ =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.     (Velocity after a period of time)

V_f  =  V_o + at

 

2.      (Distance after a period of time)

X_f = V_ot  + at²

 

3.     (Velocity/distance/acceleration without time)

V_f² =  V_o² + 2aDX

 

4.       (center seeking acceleration or centripetal acceleration)

A_c = V² / r

 

5.     (Work)

W =  D (mgh) +  D(1/2mV ²)

 

6.      (Change of momentum (Dr )=  Impulse)

 

D(mV)     =         D(f)(t)

 

7.      (Power with work)

P_avg =  W/t

 

8.      (Power with veolcity)

P = F(V)

 

9.      (Force exerted by a springy)

F_s = KX

 

10.  (Potential energy of a spring)

U_s = 1/2kx²

 

11.  (period of a spring)

T_s = 2p √ (m/k)

12.  (period of a pendulum)

T_p = 2p √ (L/g)    or    2√ L (on Earth)

13.   (relationship between period and frequency)

T = 1/f

14.  (Newton’s Law of Gravitation)

F_g = G (m₁m₂ ) / r²

 

 

15.  Pressure (with atmosphere) of a liquid

 P = p_o  + r g h

 

16.  (Force of buoyancy)

F_buoy =  weight of Liq displaced = mg = DVg

 

17.  (Equation of continuity for a liquid)

A₁V₁ =  A₂V₂

 

Test on Mon/Tues  18/19 Oct (Chapters 4-6)

                     

1.  Suppose you are whirling a ball tied to the end of

 a string in a horizontal circle and all you do is double

 its speed (tangential Velocity). Compare:

a. Its velocity______?

b. Its linear acceleration (after the speed is   increased) _____?

c. Its centripetal acceleration (after the speed is increased) _____?

d. Its centripetal force ____?

 

 

ans.

          F_c = ma_c = mV²/r

a. V increases by 2 times

b. No Change (after the speed increase)

c. a_c  increases by 4 times

d. F_c  increases by 4 times

 

 

 

2. What is the Power output of a machine that is moving a 200kg object straight up at a constant speed of 5 m/s?

 

 

 

 

 

 

 

 

 

 

 

ans.

P = W/t   or  {(F)(D)}/t   or   (F)(V) 

P  =              (N)(M)/t     or   (N)(M)/t

P  =                        (2,000N)(5m/s) = 10,000 Watts

 

 

3. A trunk is pulled across a horizontal floor at a constant speed by a 100 N force (for 2 minutes) applied by a rope at an angle of 30° above the horizontal. 

 a. The work done by friction after moving the trunk 50.0 m is:

b. Power?

 

ans.

a. W = (F)(cos30)(Dist)

    W = (100)(.87)(50)  = 4,350 J

b. P = (F)(D)/t  = 4,350J/120s  = 36.25 Watts

 

 

 

 

 

4. An object slides down a frictionless surface from rest 20m high.

a. What is its speed at the bottom?

b. Next a push starts it down at 20.0 m/s.  At what speed does it now arrive at the bottom?

 

 

 

 

 

 

 

ans. Mechnical Energy is Conserved

    a.   Gain of KE = Loss of PE    

              1/2 mV²  = mgh

          (.5)V²  = (10)(20)  --> V = 20m/s

b.   Final KE = Loss of PE + initial KE   

              1/2 mV²  = mgh  +  1/2 mV²

              (.5)V²  = (10)(20) + (.5)(20)²

                   V²  = {(200) + (200)}(2)

                V = 28.3m/s

 

 

5. The mass of a car is 150 kg.  The constant power supplied by the engine is 20 kW and the car starts from rest.

 At the end of 12 seconds, the speed of the car would be?

 

 

 

ans. (Power)(time)  = Work  = DKE + DPE

   (20,000W)(12s)  =        1/2 mV²   +    0

    (240,000J)(2)/150 = V²

       V = 56.7 m/s

 

 

 

 

Provide Pre-Test Ch 4-6