Term1 Review

Physics
Chicopee Comprehensive High School
Mr. Clark

Average Velocity:

V_avg = final position - initial position / final time - initial time = (d₂ - d₁) / (t₂ - t₁);



1. At 2:00 pm a car, traveling at a constant velocity of 94 km/hr toward the west, is 20 km to the west of Chicopee Comp. Where will it be at 4:30 pm?

   Depress mouse to view answer. Answer: d₂ = 255 km west of comp. v_avg = (d₂ - d₁) / (t₂ - t₁) t₂ - t₁ = 4:30 - 2:00 = 2.5 hr d₂ = d₁ + v_avg (t₂ - t₁) = 20 km + (94 km/hr)(2.5 h) = 255 km west of Comp.

    

    

   Acceleration:

   a_avg = chg in v / chg in t
   v_f = v_i + at
   d = 1/2(v_f + v_i)t
   d = d_i + v_it + 1/2 at²
   v_f² = v_i² + 2ad

2. A race car's velocity increases form +5 m/s to +42 m/s over a 4.5 s time period.
   What is the car's acceleration?
   

   Depress mouse to view answer. Answer: a = 8.22 m/s/s a = chg in v / chg in t a = (42 m/s - 5 m/s)/(4.5 s) a = 8.22 m/s/s

    

    

3. A car is traveling at 35 m/s. A Dog crosses the street in front of him. He steps on the brakes, coming to a stop in 4 s, just missing the dog.
   What is the average acceleration of the bus while braking?

   Depress mouse to view answer. Answer: a = - 8.75 m/s/s a = (v2 - v1) / (t2 - t1) a = (0 m/s - 35 m/s) / (4 s) a = - 8.75 m/s/s

    

    

4. A train traveling at 28 m/s slows down to 2.5 m/s with a constant acceleration of - 3.0 m/s².
   How long does this take.

   Depress mouse to view answer. Answer: t = 8.5 s vf = vi + at so t = (vf - vi) / a t = (2.5 m/s - 28 m/s) / -3 m/s/s t = 8.5 s

    

    

5. A motorcycle accelerates at a constant rate from 10 m/s to 30 m/s while going a distance of 140 m.
   How long does it take to achieve this?

   Depress mouse to view answer. Answer: t = 7.0 s d = 1/2(vf + vi)t so t = 2d / (vf + vi) t = 2(140 m)/(30 m/s + 10 m/s) t = 7.0 s

    

    

6. A drag racer completes a quarter mile run (402.5 m) in 10 s from a standing start. Assuming the acceleration was constant, calculate the acceleration in m/s².

   Depress mouse to view answer. Answer: a = 8.05 m/s^2 d = di + vi t + 1/2 at^2 vi = 0 so a = 2(d - di) / t^2 a = 2(402.5 m - 0 m) / (10 s)^2 a = 8.05 m/s^2

    

    

7. Your driving your old beat-up VW micro-bus at a velocity of 90 km/h, when rounding a corner you see a tree down across the road 50 m ahead. You hit the breaks as hard as you can and accelerate at a negative 7.5 m/s².
   How far do you go before coming to a complete stop? Can you avoid hitting the tree?

   Depress mouse to view answer. Answer: d = 42 m vf^2 = vi^2 + 2ad so d = vf^2 - vi^2 / 2a convert km/h to m/s 90 km/h x 1000 m/km x 1 h/3600s = 25 m/s d = 0^2 - (25 m/s)^2 / (2)(-7.5 m/s^2) d = 42 m You didn't hit the tree.

    

    

8. A 3 kg flowerpot falls of an apartment building porch railing 12 m above the side walk.
   With what speed does the flowerpot strike the sidewalk?

   Depress mouse to view answer. Answer: vf = 15.3 m/s vf^2 = vi^2 + 2gd so vf = sq. rt. of (vi^2 + 2gd) vf = sq. rt. of [0 + (2)(-9.8 m/s^2)(-12.0 m) vf = sq. rt. of 235.2 m^2/s^2 vf = 15.3 m/s

    

    

9. Matt is flying a light aircraft and is rising at 4 m/s when he drops his lunch bag. After 1.5 sec.: (a) what is the velocity of the bag? (b) How far has the bag fallen?(c) How far below the aircraft is the bag?

   Depress mouse to view answer. Answer: (a) vf = -10.7 m/s (b) d = -5.03 (c) 11.03 m (a) vf = vi + gt vf = 4.0 m/s + (-9.8 m/s)(1.5 s) vf = -10.7 m/s (b) d = vi t + 1/2 gt^2 d = (4.0 m/s)(1.5 s) + 1/2 (-9.8 m/s)(1.5 s)^2 d = -5.03 m (c) the aircraft has risen d = vi t (4 m/s)(1.5 s) = 6 m The bag is 5.03 m below the origin and 11.03 m below the aircraft

    

    

10. A ball is thrown up with an initial velocity of +20.0 m/s. (a) How high does the ball rise? (b) How long does the ball remain in the air?

    Depress mouse to view answer. Answer:(a) d = 20.41 m (b) 4.08 s (a) At the top of the ball's flight v = 0 so vf^2 = vi^2 + 2gd hence 2 gd = -vi^2 thus d = -vi^2 / 2g d = -(20.0 m/s)^2 / 2(-9.8 m/s^2) d = 20.41 m (b) vf = vi + gt vf = 0 so t = -vi / g t = -20.0 m/s / -9.8 m/s^2 = 2.04 s therefore 2.04 s up + 2.04 s back down = 4.08 s total time in the air.

    Forces:

    F = ma
    W = mg
    F_f = mu F_N

     

     

11. A 50 kg little red wagon is pushed with a horizontal net force of 20 N for 3.0 s. If the wagon starts from rest, how far will it go during the time the force is applied

    Depress mouse to view answer. Answer:d = 1.8 m First determine the acceleration. F = ma so a = F / m 20 N / 50 kg = 0.4 m/s/s Next find the displacement using: d = vi t + 1/2 at^2 vi = 0 so d = 1/2 at^2 d = 1/2 (0.4 m/s/s)(3.0 s)^2 d = 1.8 m

    Consider the system below for question #12:

    

     

     

12. A Horizontal force of 70 N is applied to a box with a mass of 20 kg. sitting on a table. The coefficient of friction between the box and the table is (0.30). (a) Does the box move? (b) If the box does move find it's acceleration.

    Depress mouse to view answer. Answer: (a) Yes (b) 0.56 m/s/s (a) Ff = muN N = (20 kg.)(9.8 m/s/s) = 196 N Ff = (0.30)(196 N) = 58.8 N Fa = 70 N Fnet = 70 N - 58.8 N = 11.2 N The box moves (b) F = ma a = F / m a = 11.2 N / 20 kg = 0.56 m/s/s

     

     

13. The Space Shuttle has a mass of 2.0 X 10⁶kg. At lift-off the engines generate an upward force of 30 X 10⁶N. (a) What is the weight of the shuttle? (b) What is the acceleration of the shuttle when launched? (c) The average acceleration of the shuttle during its 10 min. launch is 13 m/s/s. What velocity does it attain? (Merrill pg.102)

    Depress mouse to view answer. Answer:(a) W = 20 X 10^6 N (b) a = 5.0 m/s/s (c) v = 7.8 X 10^3 m/s (a) W = mg W = (2 X 10^6 kg)(9.8 m/s/s) = 20 X 10^6 N (b) ma = Fnet = Fappl + W Fnet = 30 X 10^6 N + (-20 X 10^6 N) Fnet = 10 X 10^6 N therefore a = Fnet / m a = 10 X 10^6 N / 2.0 X 10^6 kg a = 5.0 m/s/s (c) v = at = (13 m/s/s)(600 s) v = 7.8 X 10^3 m/s

     

    Practice Vectors in Notes

     

    

     

    Torque:

    For a sytem in equilibrium:
    ccw = cw
    £T = 0
    ccw + cw = 0
    Consider the system below for question #14:

    

14. Determine the location of the 30N Force.

    Depress mouse to view answer. Answer: x = 85.83 cm ccw = cw (35 N)(35 cm)=(30 N)(X) 1225 N*cm = 30 N*x x = 40.83 cm to the right of the fulcrum (or) x = 85.83 cm

     

     

    Consider the system below for question #15:

    

15. Determine the Force at 40 cm.

    Depress mouse to view answer. Answer: x = 25.0 N (25 N)(50 cm) + (X)(10 cm) = (30 N)(50 cm) 1250 N*cm + 10 cm*x = 1500 N*cm 10 cm*x = 250 N*cm x = 25.0 N

     

     

    Consider the system below for question #16:

    

16. Determine the Force at 80 cm.

    Depress mouse to view answer. Answer: x = 3.5 N Force (center of mass meter stick) = 0.50 N @ 50 cm (5 N)(35 cm) + (4 N)(10 cm) - (2 N)(45 cm) = (0.50 N)(5 cm) + (x)(35 cm) or (5 N)(35 cm) + (4 N)(10 cm) = (2 N)(45 cm) + (0.50 N)(5 cm) + (x)(35 cm) 175 N*cm + 40 N*cm = 90 N*cm + 2.5 N*cm + 35 cm*x 215 N*cm = 92.5 N*cm + 35 cm*x 122.5 N*cm = 35 cm*x x = 3.5 N

     

     

    Consider the system below for question #17:

    

17. Determine the position of the fulcrum to put the system into equilibrium
    

    Click to see answer

     

     

    Consider the system below for question #18:

    

18. If A = 3N, determine the weight of B,C,D,E, and F
    (If you can do this, you understand torque)
    

    Click to see answer

     

     

    Depress mouse to view answer. Answer:

    TO BE CONTINUED ???