College Physics for AP® Courses

# Chapter 7

### Problems & Exercises

1.

3.

(a) $5.92×105J5.92×105J size 12{5 "." "92" times "10" rSup { size 8{5} } " J"} {}$

(b) $−5.88×105J−5.88×105J size 12{ - 5 "." "88" times "10" rSup { size 8{5} } " J"} {}$

(c) The net force is zero.

5.

$3 . 14 × 10 3 J 3 . 14 × 10 3 J size 12{3 "." "14" times "10" rSup { size 8{3} } " J"} {}$

7.

(a) $− 700 J − 700 J size 12{ - "700"J} {}$

(b) 0

(c) 700 J

(d) 38.6 N

(e) 0

9.

$1 / 250 1 / 250 size 12{1/"250"} {}$

11.

$1 . 1 × 10 10 J 1 . 1 × 10 10 J size 12{1 "." 1 times "10" rSup { size 8{"10"} } " J"} {}$

13.

$2 . 8 × 10 3 N 2 . 8 × 10 3 N size 12{2 "." 8 times "10" rSup { size 8{3} } " N"} {}$

15.

102 N

16.

(a) $1.96 × 1016 J1.96 × 1016 J size 12{1 "." "96" times "10" rSup { size 8{"16"} } " J"} {}$

(b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb.

18.

(a) 1.8 J

(b) 8.6 J

20.

$v f = 2 gh + v 0 2 = 2 ( 9.80 m /s 2 ) ( − 0 .180 m ) + ( 2 .00 m/s ) 2 = 0 .687 m/s v f = 2 gh + v 0 2 = 2 ( 9.80 m /s 2 ) ( − 0 .180 m ) + ( 2 .00 m/s ) 2 = 0 .687 m/s size 12{v rSub { size 8{f} } = sqrt {2 ital "gh"+v rSub { size 8{0} rSup { size 8{2} } } } = sqrt {2 $$9 "." "80"" m/s" rSup { size 8{2} }$$ $$- 0 "." "180"" m"$$ + $$2 "." "00 m/s"$$ rSup { size 8{2} } } =0 "." "687"" m/s"} {}$

22.

$7.81 × 10 5 N/m 7.81 × 10 5 N/m size 12{7 "." "81" times "10" rSup { size 8{5} } " N/m"} {}$

24.

9.46 m/s

26.

27.

Equating $ΔPEgΔPEg size 12{Δ"PE" rSub { size 8{g} } } {}$ and $ΔKEΔKE size 12{Δ"KE"} {}$, we obtain $v=2gh + v02=2(9.80 m/s2)(20.0 m)+(15.0 m/s)2=24.8 m/sv=2gh + v02=2(9.80 m/s2)(20.0 m)+(15.0 m/s)2=24.8 m/s size 12{v= sqrt {2 ital "gh"+v rSub { size 8{0} rSup { size 8{2} } } } = sqrt {2 $$9 "." "80"" m/s" rSup { size 8{2} }$$ $$"20" "." 0" m"$$ + $$"15" "." "0 m/s"$$ rSup { size 8{2} } } ="24" "." 8" m/s"} {}$

29.

(a) $25×106years25×106years size 12{"25" times "10" rSup { size 8{6} } "years"} {}$

(b) This is much, much longer than human time scales.

30.

$2 × 10 − 10 2 × 10 − 10 size 12{2 times "10" rSup { size 8{-"10"} } } {}$

32.

(a) 40

(b) 8 million

34.

\$149

36.

(a) $208 W208 W$

(b) 141 s

38.

(a) 3.20 s

(b) 4.04 s

40.

(a) $9.46×107 J9.46×107 J size 12{9 "." "46" times "10" rSup { size 8{7} } " J"} {}$

(b) $2.54 y2.54 y size 12{2 "." "54" times "10" rSup { size 8{7} } " J"} {}$

42.

Identify knowns: $m=950 kgm=950 kg size 12{m="950""kg"} {}$, $slope angleθ=2.00ºslope angleθ=2.00º$, $v=3.00 m/sv=3.00 m/s$, $f=600 Nf=600 N size 12{f="600"N} {}$

Identify unknowns: power $PP size 12{P} {}$ of the car, force $FF size 12{F} {}$ that car applies to road

Solve for unknown:

$P = W t = Fd t = F d t = Fv , P = W t = Fd t = F d t = Fv , size 12{P= { {W} over {t} } = { { ital "Fd"} over {t} } =F left ( { {d} over {t} } right )= ital "Fv",} {}$

where $FF size 12{F} {}$ is parallel to the incline and must oppose the resistive forces and the force of gravity:

$F = f + w = 600 N + mg sin θ F = f + w = 600 N + mg sin θ size 12{F=f+w="600 N"+ ital "mg""sin"θ} {}$

Insert this into the expression for power and solve:

P = f + mg sin θ v = 600 N + 950 kg 9.80 m/s 2 sin 2º ( 30.0 m/s ) = 2.77 × 10 4 W P = f + mg sin θ v = 600 N + 950 kg 9.80 m/s 2 sin 2º ( 30.0 m/s ) = 2.77 × 10 4 W alignl { stack { size 12{P= left (f+ ital "mg""sin"θ right )v} {} # size 12{ {}= left ["600 N"+ left ("950 kg" right ) left (9 "." "80 m/s" rSup { size 8{2} } right )"sin2"° right ] $$"30" "." "0 m/s"$$ } {} # =2 "." "77" times "10" rSup { size 8{4} } W {} } } {}

About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline.

44.

(a) 9.5 min

(b) 69 flights of stairs

46.

641 W, 0.860 hp

48.

31 g

50.

14.3%

52.

(a)

(b)

(c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b)

54.

(a) 108 kJ

(b) 599 W

56.

(a) 144 J

(b) 288 W

58.

(a)

(b) 2.52%

(c) (14 metric tons)

60.

(a) 294 N

(b) 118 J

(c) 49.0 W

62.

(a) $0.500 m /s20.500 m /s2 size 12{0 "." "500"" m/s" rSup { size 8{2} } } {}$

(b) $62.5 N62.5 N$

(c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since $f=F−maf=F−ma size 12{f=F - ital "ma"} {}$. If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared ($t2t2 size 12{t rSup { size 8{2} } } {}$). Therefore, the water resistance will not depend linearly on the velocity.

64.

(a) $16.1×103N16.1×103N size 12{"16" "." 1 times "10" rSup { size 8{3} } N} {}$

(b) $3.22×105J3.22×105J size 12{3 "." "22" times "10" rSup { size 8{5} } J} {}$

(c) 5.66 m/s

(d) 4.00 kJ

66.

(a) $4.65×103 kcal4.65×103 kcal size 12{4 "." "65" times "10" rSup { size 8{3} } " kcal"} {}$

(b) 38.8 kcal/min

(c) This power output is higher than the highest value on Table 7.5, which is about 35 kcal/min (corresponding to 2415 watts) for sprinting.

(d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!).

69.

(a) 4.32 m/s

(b) $3.47×103N3.47×103N size 12{3 "." "47" times "10" rSup { size 8{3} } N} {}$

(c) 8.93 kW

### Test Prep for AP® Courses

1.

(b)

3.

(d)

5.

(a)

7.

The kinetic energy should change in the form of –cos, with an initial value of 0 or slightly above, and ending at the same level.

9.

Any force acting perpendicular will have no effect on kinetic energy. Obvious examples are gravity and the normal force, but others include wind directly from the side and rain or other precipitation falling straight down.

11.

Note that the wind is pushing from behind and one side, so your KE will increase. The net force has components of 1400 N in the direction of travel and 212 N perpendicular to the direction of travel. So the net force is 1420 N at 8.5 degrees from the direction of travel.

13.

Gravity has a component perpendicular to the cannon (and to displacement, so it is irrelevant) and has a component parallel to the cannon. The latter is equal to 9.8 N. Thus the net force in the direction of the displacement is 45 N − 9.8 N, and the kinetic energy is 53 J.

15.

The potato cannon (and many other projectile launchers) above is an option, with a force launching the projectile, friction, potentially gravity depending on the direction it is pointed, etc. A drag (or other) car accelerating is another possibility.

17.

The kinetic energy of the rear wagon increases. The front wagon does not, until the rear wagon collides with it. The total system may be treated by its center of mass, halfway between the wagons, and its energy increases by the same amount as the sum of the two individual wagons.

19.

(d)

21.

0.049 J; 0.041 m, 0.25 m

23.

20 m high, 20 m/s.

25.

(a)

27.

(d)

29.

(c)

31.

(b)

33.

(c)

35.

(c)

37.

(c), (d)

39.

(a)

41.

(b)

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