Download Newton`s First Two Laws

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Honors Physics, Chunk Five, Applications of Newton’s Laws, Worksheet (key)
Friction
1. (a) What causes friction between two solid surfaces? (b) After the roughness of the surfaces, what most affects the
frictional force experienced? (a) Friction is caused by irregularities on the surfaces of the two objects that are
touching. (b) How much the surfaces are pressed together
2. (a) What causes fluid friction? (b) What two factors affect the amount of fluid friction experienced? (a) Fluid friction
is caused by the interaction between the fluid and the object passing through it. (b) How fast the object is
moving through the fluid and the type of fluid the object is moving through (i.e. water or air).
3. (a) What is the net force on a 10 N falling object that encounters 4 N of air resistance? (b) 10 N of air resistance?
(a) 6 N down (b) zero
4. What is the acceleration of a falling object that has reached its terminal velocity? zero
5. Why does a heavy skydiver fall faster than a lighter one who wears the same size parachute? This is referring to after
the parachute is opened. The reason the heavy skydiver falls faster than the lighter one after their parachutes
are opened is that the heavier skydiver takes longer to reach terminal velocity than the lighter one does.
Therefore, the heavier skydiver’s terminal velocity is faster than the lighter skydiver’s terminal velocity.
6. Is a skydiver who has reached his terminal speed in freefall? Explain No. Because the only force acting on an
object in free fall is gravity. In this case, the object has both the forces of gravity and friction acting on it.
7. (a) How does the weight of a falling body compare with the air resistance it encounters before it reaches terminal
velocity? (b) After? (a) The weight of a falling body is greater than its air resistance before it reaches terminal
velocity. (b) After it reaches terminal velocity, the weight of the object is equal to its air resistance.
8. Why is it that a cat that falls from the top of a 50 story building will hit the ground at the same speed as it would if it fell
from the 20th story? Both cats would have reached terminal velocity.
9. If Galileo dropped two balls from the top of the Leaning Tower of Pisa, air resistance was not really negligible.
Assuming the balls were the same size and shape, one made of wood and the other of metal, which ball struck the
ground first? Explain. The metal ball would hit the ground first because the wooden ball weighed less and
would, therefore, be influenced by air resistance more than the metal ball would. If they both reached terminal
velocity, the wooden ball would reach it before the metal ball would.
10. What will be the acceleration of a skydiver when air resistance builds up to be half his weight? 4.9 m/s2
Pressure
11. A stack of 5 physics books is lying on a table. Each book has a mass of 1.5 kg. The dimensions of the back cover of
each book is 22 cm x 26.5 cm. Neglect the effects of air pressure. g = 10 m/s2
a. What is the force exerted on the table by one book? Fw = m g = (1.5 kg)(10 m/s2) = 15 N = Fw
b. What is the area of the back cover of each book in m 2? (0.22 m)(0.265 m) = 0.0583 m2 = area of the back cover
c. What is the pressure exerted by one book on the table? P = F / A = 15 N / 0.0583 m2 = 257.3 Pa = P
d. What is the force exerted on the table by the stack of books. (15 N)(5 books) = 75 N = F
e. What is the pressure exerted by the stack of books on the table? P = F / A = 75 N / 0.0583 m2 = 1286.4 Pa = P
12. A certain physics student weighs 650 N and she standing with both feet on the floor. If the total pressure exerted on
the ground by her is 65 N/m2, what is the total area of her feet contacting the ground? P = F / A A = F / P = 10 m2
Gravitation
13. Write the Newton’s Law of Universal Gravity equation, defining all terms: Fgravity = G m1 m2 / d2
The force of gravity (Newtons) = the gravitation constant (N m 2 / kg2) times the mass of one object (kg) times
the mass of the other object (kg) divided by the square of the distance between the two objects (m2)
14. Complete: Gravity force is directly proportional to the product of the masses and inversely proportional to the
square of the distance between them.
15. If the gravitational force of attraction between two objects is 100N, what would it be if the distance between them were
(a) doubled (b) halved (c) tripled?
(a) 25 N (b) 400 N (c) 11.1 N
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16. If the mass of two objects is doubled but the distance between them remains the same, what happens to the force of
gravitational attraction between them? The gravitational force between them is multiplied by 4.
17. What is the force of gravity between a 50 kg student and a 60 kg student 2 meters apart?
Fgrav = G (50 kg) (60 kg) / 22 = 750 G
Circular Motion
18. (a) What is the force that acts on an object in a circular motion? (b) In what direction does this force act?
(a) centripetal force (b) This force always acts toward the center of the circle.
19. Why do you feel like you are flung sideways when your car travels around a sharp curve? You feel this because of
Newton’s first law, the law of inertia. As long as your car travels around this curve, the tires on the road are
causing the centripetal force in the direction of the center of the circle that the curve is a part of. If you
suddenly drove straight, the tendency to slide sideways in the car would disappear because there would no
longer be a centripetal force acting on the car.
20. Swing a bucket of water around in a full circle. (a) Does the water stay in the bucket? (b) Explain your choice. (a) yes
(b) If you swing the bucket fast enough, although the water still falls, it won’t fall out of the bucket. The reason
for this is that, if the bucket is moving fast enough, it will fall as fast as the water does and the water will stay
in the bucket. This is similar to a satellite orbiting, or “falling around” the earth.
21. When you observe an object moving in a circle, what can you infer about the net force acting on it? You can infer
that the net force acting on it is directed toward the center of this circle.
22. What holds the earth in its orbit around the sun? The gravitational force of the sun acting on the earth is the
centripetal force pulling the earth toward the center of the sun. If the earth didn’t have sufficient tangential
velocity, it would fall into the sun.
23. Why must there be a force acting for an object to successfully make it around a curved path? This force acting
toward the center of the circle keeps the object in the curved path.
24. (a) What causes the force on a car as it travels through a curve? (b) Why does a car skid on an icy curve? (a) The
friction between the tires and the road provides the centripetal force that keeps the car on the road while it is
in the curve. (b) The car skids because the friction force is significantly decreased on an icy road. Therefore
the centripetal force is decreased and the car cannot stay in the curve.
25. Listed below are a number of situations involving circular motion. In each case, identify the source of the force
needed to keep the objects in question moving in their circular paths.
Example: A race car going around a corner: Friction between the road and the car’s tires provides the centripetal
force that holds the car in a circular path.
a) a child riding on a merry-go-round: The beam holding the horse to the center of the ride provides the
centripetal force that holds the child in a circular path.
b) a ball at the end of a string being swung in a horizontal circle: The string provides the centripetal force that
holds the ball in the circular path.
c) a sprinter running around the curve at the end of the track: Friction between track and shoes provides the
centripetal force that holds the sprinter in a circular path.
d) you in a car as you go over the top of a hill: Gravity between you and the earth provides the centripetal force
to hold you in the curve as you go over the hill (provided that you do not go too fast).
e) you in your seat on a roller coaster going through the bottom of a dip: The bottom of the car you are sitting in
provides the centripetal force to hold you in the circular path.
f) a little girl being swung around in a horizontal circle by her daddy: The little girl’s arms and her daddy’s arms
provide the centripetal force that keeps the little girl in the circular path.
g) you in a car going around a horizontal corner: Friction between you and the car seat provides the centripetal
force that keeps you moving in the circle as long as you are traveling around the corner.
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h) you in the "Gravitron", a carnival ride where you stand inside a spinning drum and are pressed against the side:
The wall of the ride provides the centripetal force that keeps you moving in a circle as long as the ride is
turned on.
i) Mud sticking in the tread of a spinning automobile tire: Friction between the mud and the tire provides the
centripetal force that keeps the mud on the spinning tire.
j) you turning on roller skates or roller blades: Friction between the roller blade tires and the road provides the
centripetal force that causes you to stay in your turn (circle).
k) water in a bucket being swung in a vertical circle: The bucket handle and the person’s arm and the bottom of
the bucket provide the centripetal force that allows the water to travel in a circle and not fall out of the bucket.
l) Mars going around the sun: The gravitational force between Mars and the sun provides the centripetal force
that causes Mars to orbit the sun.
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1. most
2. a, b
3. d, e
4. still downward
5. c, f
6. c, f
7. c
8. greater
(a) a = 10 m/s2
(b) a = 6 m/s2
(c) a = 0 m/s2
(d) a = –2 m/s2
(e) a = –10 m/s2
(f) a = 0 m/s2
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Centripetal Force Problems
1. A roller coaster car speeds down a hill past point A. It then rolls up a hill past point B as shown to the right. If the car
has a tangential speed of 20 m/s at point A and the track exerts a centripetal force of 2.06 x 104 N on the car at point
A, what is the car’s mass?
vT (pt A) = 20 m/s, Fc = 2.06 x 104 N, r1 = 10 m,
r2 = 15 m, g = 9.8 m/s2, mcar = ?
Fc = mcar vT2 / r
radius = 10 m
radius = 15 m
B
mcar = Fc r / vT2
mcar = (2.06 x 104 N) (10 m) / (20 m/s)2 = 515 kg = mcar
A
2. Tarzan tries to cross a river by swinging from one bank to the other on a vine that is 10 m long. His speed at the bottom
of the swing, just as he clears the river’s surface, is 8 m/s. Tarzan doesn’t know that the vine has a breaking strength of
1,000 N. What is the largest mass that Tarzan can have to make it safely across the river without the vine breaking?
free body diagram
r = 10 m, vT = 8 m/s, Frope max = 1,000 N, g = 9.8 m/s2, mmax = ?
FC
Ftotal on rope = Fc + Fw
Ftotal on rope ≤ Frope max
Fw = m g
Fc = m vT2 / r
Frope max = (mmax) (vT2) / r + (mmax) (g)
Frope max = (mmax) (vT2 / r + g)
mmax = Frope max / (vT2 / r + g)
mmax = 1,000 N / (8 m/s)2 / 10 m + 9.8 m/s2
mmax = 62 kg
FW
3. A 13,500 N car traveling at 50 km/h rounds a curve of radius 2 x 10 2 m.
a. What is the centripetal acceleration of the car?
b. What is the force that maintains centripetal acceleration?
c. What is the minimum coefficient of friction between the tires and the road that will allow the car to round the curve
safely (without skidding)?
FW = 13,500 N, r = 200 m, vT = 50 km/hr, g = 9.8 m/s2, aC = ?, FC = ?, µ = ?
a.
aC = vT2 / r = [(50 km / hr) (1,000 m / km) (1 hr / 3,600 s)]2 / 200 m = 0.965 m/s2 = aC
b.
FC = m aC
FW = m g
m = FW / g
FC = (FW / g) aC
FC = (13,500 N / 9.8 m/s2) (0.965 m/s2) = 1,329 N = FC
c.
F f = µ FN
assume: FC = Ff (if car is to just stay in the curve without skidding)
µ = Ff / FN (don’t know any of these, so must substitute)
µ = FC / FW = 1,329 N / 13,500 N = 0.098 = µ
F N = FW
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4. A 2 x 103 kg car rounds a circular turn of radius 20 m. If the road is flat and the coefficient of friction between the tires
and the road is 0.70, how fast can the car go without skidding?
m = 2,000 kg, r = 20 m, µ = 0.70, g = 9.8 m/s2, vT(without skidding) = ?
FW = m g = F N
Ff = µ F N
FC = m vT2 / r
vT = √ µ g r
F C = m v T2 / r
Ff = m vT2 / r
= √(0.70) (9.8 m/s2) (20 m)
FC = Ff (at just the point before skidding)
µ FN = m vT2 / r
µ m g = m vT2 / r
= 11.7 m/s = vT
Gravitational Force Problems
5. Mars has a mass of 6.4 x 1023 kg and its moon, Phobos, has a mass of 9.6 x 1015 kg. If the gravitational force
between the two bodies is 4.6 x 1015 N, how far apart are Mars and Phobos?
mMars = 6.4 x 1023 kg, mPhobos = 9.6 x 1015 kg, FG = 4.6 x 1015 N, G = 6.67 x 10-11 N m2 / kg2
dMars to Phobos = ?
FG = (G) (mMars) (mPhobos) / (dMars to Phobos)2
dMars to Phobos = √(G) (mMars) (mPhobos) / FG
dMars to Phobos = √(6.67 x 10-11 N m2 / kg2) (6.4 x 1023 kg) (9.6 x 1015 kg) / 4.6 x 1015 N
= 9.44 x 106 m
6. The earth has a mass of 5.98 x 1024 kg and the sun has a mass of 1.99 x 1030 kg. If the distance between the earth
and the sun is 1.5 x 108 km, what is the gravitational force between the earth and the sun?
mearth = 5.98 x 1024 kg, msun = 1.99 x 1030 kg, dearth to sun = 1.5 x 1011 m, G =6.67 x 10-11 N m2 / kg2,
FG between sun and earth = ?
FG = (G) (mearth) (msun) / (dearth to sun)2 = (6.67 x 10-11 N m2 / kg2) (5.98 x 1024 kg) (1.99 x 1030 kg) / (1.5 x 1011 m)2
FG = 3.53 x 1022 N