Practice Problems for 1st Semester Exam

Practice Problems for 1 st Semester Exam
1. How far would you travel if you maintain a velocity of 10 m/s for 30 s?
v = 10 m/s t = 30 s v = d/t d =vt d = (10 m/s) ( 30s) d = 300m
2. Brendan is driving a go-kart at 30 m/s. He slams on the brakes and comes to a stop in 3s.
What was the acceleration of Armando’s go-kart?
vo = 30 m/s vf = 0 m/s t = 3s a=∆v a = (vf - vo) /t a = (0m/s – 30m/s) / 3s a = - 10 m/s
t
3. While on a rollercoaster at Busch Gardens, Akisha’s sandal falls off.
a) If there were no air resistance, what would be the velocity of the sandal after falling for 5.5s?
t = 5.5s v = ? v = gt v = (10 m/s 2 )(5.5s) v = 55 m/s
b) How far would the sandal fall during this time?
d=1/2 gt 2 d = ½ (10 m/s 2 ) ( 5.5s) 2 d = ½ (10 m/s 2 ) (30.25 s 2 ) d = 151.25 m
4. You drop an egg from the bleachers that are 9.5m high.
a) What will be the reading on a stop watch after recording the elapsed time of the egg’s fall
d = 9.5m t = 2d /g t = 2(9.5m)/ (10 m/s 2 ) t = 1.9 s 2 t = 1.38 s
b.) What will be the velocity of the egg just before it hits the ground?
t = 1.38 s v = ? v = gt v = (10 m/s 2 )(1.38s) v = 13.8 m/s
5. Nicole is flying an airplane north at 100km/ h and has a easterly crosswind of 50 km/ h.
What is the resulting velocity of her plane?
100 km/h 100km/h 100 km/h v vy = 100 km/h
50 km/h 50 km/h vx = 50 km/h
v 2 = vx 2 + vy 2 v 2 = (50 km/h) 2 + (100 km/h) 2
v 2 = (2,500 km 2 /h 2 ) + (10,000 km 2 /h 2 )
v 2 = 12,500 km 2 /h 2
v 2 = 12,500 km 2 /h 2 v = 111.80 km/h

6. Troy, a Pinellas county sheriff is trying to determine the speed of a car that went off the Howard Franklin
bridge on a rainy night and landed in the water 5 m below. The car landed 10 m horizontally from the
bridge in the water. How fast was the car going when it left the road?
PS: the driver was not hurt , however Troy gave him a ticket for reckless driving!
dy = 5m dx = 10m vx = ? vx = dx / t don’t know t; solve for t first
t = 2dy / g t = 2(5m) /(10 m/s 2 ) t = 10m/ 10 m/s 2 t = 1s
vx = dx / t vx = (10m) / 1s vx = 10 m/s
7. Mrs. Peck’s class is trying to calculate where a flour sack would land if dropped from a moving airplane.
The plane would be moving horizontally at a constant speed of 50 m/ s at an altitude of 200m. Neglecting
air resistance, how far horizontally from the dropping point would they predict the flour sack would land?
vx =50 m/s dy = 200m dx = ? dx = vx t don’t know t; solve for t first
t = 2dy / g t = 2(200m)/10 m/s 2 t = 400m /10 m/s 2 t = 40 s 2 t = 6.32 s
dx = (50 m/s) (6.32 s) dx = 316 m
8. A pull of 6N causes a 0.5 kg object to move at a constant velocity. Complete the diagram.
a. Fn = 5N b.
Ff = c. 6N constant velocity Fapp = Ff & Fnet = 0N
d. Fapp = 6N
m = 5kg … Fg = mg … Fg = (.5kg)(10m/s
Fg e. =
f.
m = _______
g.
h.
2 ) = 5N
5N
.5 kg Fg = Fn … Fn = 5N
0N a = Fnet/ m a = 0N/ .5kg = 0 m/s 2
Ff =
c.
a = _______
a. Fn = b.
Fg e. =
d. Fapp =
f.
m = _______ 8 kg Fnet = Fapp – Ff = 30N – 10N = 20N
g.
h.
0 m/s 2
9. An applied force of 30N is used to accelerate an object with a weight of 80N to the right across
a frictional (with friction) surface. The object encounters 10N of friction. Complete the diagram.
10N
a = _______
80N Fg = Fn Fn = 80N
80N
30N Fg = mg m = Fg/ g = 80N/ 10 m/s 2 = 8kg
20N a = Fnet / m a = 20N/ 8kg a = 2.5 m/s 2
2.5 m/s 2

10. A rightward force of 25N is applied to a 5kg object to move it across a rough surface with a
rightward acceleration of 2 m/s
a. Fn = b.
Ff = c.
2 . Complete the diagram
50N Fg = mg Fg = (5kg)(10 m/s 2 )
15N
25N Fg = Fn … Fn = 50N
Fg e. =
d. Fapp =
f.
m = _______ 5 kg Fnet = Ff + Fapp Ff = Fapp – Fn = 25N – 10N = 15N
g.
h.
10N
2 m/s 2
a = _______
50N Fnet = ma Fnet = (5kg)(2 m/s 2 ) = 10N
11. a. What is the momentum of a 55 kg carton that slides at 4 m/ s across an icy surface?
m = 55kg v = 4 m/s p = ? p = mv p = (55kg) (4m/s) p = 220 kg m/s
b. The carton skids onto a rough surface and stops in 4 s. What is the change in momentum of the carton?
t = 4s vf = 0m/s vo = 4m/s ∆p = m∆v ∆p = (55kg) (0m/s – 4m/s) ∆p = - 220kg m/s
c. What was the impulse that changed the momentum of the carton, thus bringing it to a stop
∆p = J ∆p = - 220 kg m/s J = - 220 Ns
d. Calculate the force of friction the carton encountered in order to stop it in 4 s.
J = Ft F = J/t F = -220 Ns/ 4s F = - 55N
12. a. What impulse occurs when an average force of 20 N is exerted on a cart for 5s?
F = 20N t = 5s J = ? J = Ft J = (20N) (5s) J = 100Ns
b. What change in momentum does the cart undergo?
∆p = J ∆p = 100 kg m/s
c. If the mass of the cart is 5 kg and the cart is initially at rest, calculate its final velocity?
∆p = m∆v ∆p = m (vf - vo) 100kg m/s = (5kg) ( vf – 0m/s)
100kg m/s = 5kg (vf)
100kg m/s / 5 kg = 5kg (vf) / 5 kg
20 m/s = vf
13. Hamlet, a hamster, runs on his exercise wheel, which turns around once every 3.5 s.
What is the frequency and period of the wheel?
T = 3.5s f = 1/ T f = 1/ 3.5s f = .29 Hz

14. Brian’s favorite ride at Busch Gardens is the rotor, which has a radius of 4 m.
The ride takes 2 seconds to make one full revolution. Brian’s mass is 60 kg.
r = 4m m = 60kg T = 3s
a. What is Brian’s frequency?
f = 1/T f = 1/ 3s f = .33 Hz
b. What is the period of Brian’s circular motion?
T = 3 s
c. What is Brian’s rotational velocity?
rotational velocity = frequency rotational velocity = .33 turns per second = .33Hz
d. What is Brian’s tangential velocity.
v = 2π r / T v = 2 (3.14)( 4m) / 3s v = 8.37 m/s
e. What is Brian’s centripetal acceleration?
ac = v 2 /r ac = (8.37 m/s) 2 / 4m ac = (70.0569 m 2 /s 2 ) / 4m ac = 17.51 m/s 2
f. What is the centripetal force that is exerted on Brian to maintain his circular motion?
Fc = mac Fc = (60kg) ( 17.51 m/s 2 ) F = 1050.60 N
15. Cody twirls a round piece of pizza dough overhead. It revolves .25 times in one second.
A 0.02 kg piece of pepperoni is stuck on the dough 0.13 m from the pizza’s center.
f = .25 Hz m = .02 kg r = .13 m
a.) What is the frequency of the pepperoni?
f = .25 Hz
b.) What is the pepperoni’s period?
T = 1/f T = 1/.25Hz T = 4s
c.) What is the pepperoni’s tangential speed?
v = 2π r / T v = 2 (3.14)( .13m) / 4s v = .20 m/s
d.) What is the pepperoni’s centripetal acceleration?
ac = v 2 /r ac = (.20 m/s) 2 / .13m ac = (.04 m 2 /s 2 ) / .13m ac = .31 m/s 2
e.) What is the centripetal force on the pepperoni?
Fc = mac Fc = (.02 kg) (.31m/s 2 ) F = .006 N

16. a. How much work is done on a 5 kg backpack filled with books that you carry up a flight of stairs
that is 4 m high?
m = 5 kg d = 4m W = Fd Fg = mg = (5 kg) (10 m/s 2 ) = 50N
W = Fd W = (50N) (4m) W = 200 J
b. What power is expended if you lift the backpack a distance of 4 m in a time of 10 s?
P = W/t P = 200J / 10s P = 20 watts
c. Are you expending more or less power if you lift the same backpack up the same flight of stairs
in half the amount of time ( 5 seconds ).? ____more_____________ (more or less)
How much more or less power? __________twice______________ as much power
half, twice, one fourth, 4X, one third, 3X
d. What is the gravitational potential energy of the backpack on top of the stairs?
PEg = Work is took to move it to this elevated position PEg = 200J
or calculate mathematically PEg = mgh PEg = (5kg)(10 m/s 2 )(4m) PEg =200J
17. a. Calculate the work needed to lift a 90N block of ice a vertical distance of 3m.
Fg = 90N d = 3m W = ? W = Fd W = (90N)(3m) W = 270J
b. What PE does it have?
PEg = Work is took to move it to this elevated position PEg = 270J
or calculate mathematically PEg = mgh PEg = (90N)(3m) PEg =270J
18. Calculate the potential energy of 8 million kg of water dropping over 50m high Niagara Falls.
m = 8,000,000 kg h = 50m PE = ?
PE = mgh PE = (8,000,000 kg) (10 m/s 2 ) (50m) PE = 4,000,000,000 J
19. a. Calculate the kinetic energy of a 3kg toy cart that moves at 4 m/s.
m = 3 kg v = 4 m/s KE = ? KE = ½ mv 2
KE = ½ (3kg)(4 m/s) 2 KE = ½ (3kg)(16 m 2 /s 2 ) KE = 24J
b. Calculate the kinetic energy of the same cart at twice the velocity.
double velocity … KE is quadrupled 4(24J) = 96J = KE

20. a. How much work is done on a 10 kg backpack that you carry up a flight of stairs that is 3 m high?
m = 10kg d = 3m W = ? W = Fd F = Fg Fg = mg Fg = (10kg)(10m/s 2 ) = 100N
W = Fd W = (100N)(3m) W = 300J
b. What power is expended if you lift the backpack a distance of 3 m in a time of 8 s?
t = 8s P=? P = W/t P = 300J / 8s P = 37.5 watts
c. How much power is expended lifting the same backpack up the stairs in half of the time (twice as fast)?
decrease time … increase power P ~ 1/t ½ time … double power 2(37.5 watts) = 75watts = P
mathematically solved: P = W/t P = 300J/ 4s P = 75 watts
d. What is the PEg of the backpack at the top of the stairs? (hint: you do not have to do any math on paper)
the work (changing backpack’s energy) done on the backpack is = to its Peg at top of stairs
W = 300J … PEg = 300J
mathematically solved: PEg = mgh PEg = (10kg)(10 m/s 2 )(3m) PEg= 300J