Fisica

CHAPTER 5

NEWTON’S LAWS OF MOTION

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored.
5-1.

m = 975 lb ×

1kg = 442 kg. 2.205 lb

5-2.

matom = m p + me = 1.673 × 10−27 kg + 9.11 × 10−31 kg = 1.67 × 10−27 kg. To three significant digits the mass of the electron isnegligible compared to the mass of the proton. 1.0 kg = 6.0 × 1026 atoms. N = 1.67 × 10−27 kg/atom

5-3.

matom = 8(m p + mn + me ) = 8(1.673 × 10−27 kg + 1.675 × 10−27 kg + 9.11 × 10−31 kg) = 2.68 × 10−26 kg. N = 1.0 kg = 3.7 × 1025 atoms. 2.68 × 10−26 kg/atom
mboy aboy a girl
50 kg × 7 m/s 2 = 43 kg. 8.2 m/s 2

5-4.

The forces must have equal magnitudes. Thus mboy aboy = mgirl agirl .mgirl = =

†5-5.

a =

v2 − v1 ( 80 km/h − 0 km/h ) 1000 m 1h × × = = 3.83 m/s 2 . Final answer ∆t 5.8 s km 3600 s

5-6. 5-7. 5-8.

= 3.8 m/s 2 . To find the magnitude of the average force, use the three-digit intermediate result for m a : F = ma = 1620 kg × 3.83 2 = 6205 N. Final answer = 6.2 × 103 N. s a = F/m = 2.7 × 105 N/1.6 × 104 kg = 17 m / s 2

∆v = 50 km/h − 0 km/h = 50 km/hr= 13.9 m/s F = m a = m∆v/∆t = 57 kg(13.9 m/s)/0.12 s = 6.6 × 103 N, which is 12 times the weight.
m = 16 000 metric tons × 103 kg/metric ton = 1.6 × 107 kg. F 6.7 × 105 N a = = . a = 0.0419 m/s 2 . Final result = 0.042 m/s2. 7 m 1.6 × 10 kg v − v0 50 km/h − 0 m 1h = = 332 s. Final result: t = 5.5 min. × 1000 × v = v0 + at ⇒ t = 2 0.0419 m/s km 3600 s a

†5-9.

Vector note: “Decelerates” is anontechnical way of stating that the acceleration points in the opposite direction from the velocity, causing the speed to decrease. If we take the positive direction for vectors to be in the direction of motion, then a must be represented by a negative number, and the net force will also be a negative number. A “free-body” diagram is shown to illustrate these concepts. Thus Fnet = ma = 1500 kg ×(−8.0 m/s 2 ) = −1.2 × 103 N. Again, the − sign means the force points in the opposite direction from the original motion.

Fnet Direction of motion

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5
2

5-10.

2 0 − (13.9 m/s ) v 2 − v0 v = v + 2ax. v0 = 50 km/h = 13.9 m/s , v = 0 ⇒ a = = 2x 2(0.40 m)
2 2 0

= − 241 m/s 2 . (Vector note: The − sign means the acceleration is in the opposite direction from themotion.) The magnitude of the net force is F = m a = (1400 kg)(241 m/s 2 ) = 3.4 × 105 N. 5-11.
2 v0 = 174 km/h = 48.3 m/s. Since v2 − v 0 = 2a(x − x0), 2 −v0 −(48.3 m / s) 2 = −1.8 × 103 m / s 2 a= = 2 × 0.66 m 2( x − x0 )

F = ma = −75 kg × 1.8 × 103 m/s2 = −1.3 × 105 N 5-12. 5-13. 5-14. We calculated that it had an acceleration = 3.26 × 104 m/s2 F = ma = 45 kg × 3.26 × 104 m/s2 = 1.46 × 106 N. ∆v30 m/s − 0 m/s = = 500 m/s 2 . F = ma ⇒ F = 0.07 kg × 500 m/s 2 = 35 N ∆t 0.060 s ∆v 22 m/s − 0 m/s a = = = 110 m/s 2 . F = ma = 0.035 kg × 110 m/s 2 = 3.9 N. Vector note: ∆t 0.20 s a = Becasuse both a and F are positive, they point in the same direction, as required by Newton’s Second Law. ∆v 10.0 m/s − 15.0 m/s m = − 4.17 m/s 2 . Final result = −4.2 2 . Second a = . First interval: a1 = ∆t 1.2 ss 5.0 m/s − 10.0 m/s interval: a2 = = − 2.38 m/s 2 . Final result = −2.4m/s 2 . During the first 2.1 s
interval, F1 = ma1 = 240 kg × (−4.17 m/s 2 ) = −1.0 × 103 N. During the second interval,

†5-15.

F2 = ma2 = 240 kg × (−2.38 m/s 2 ) = −5.7 × 102 N. Vector note: The − signs imply that the
5-16. accelerations and forces point in the opposite direction from the motion. F 55 N a = = = 2037m/s 2 . v final = vinitial + a ∆t = 0 + 2037 m/s 2 × 0.13 s = 264.8m/s. m 0.057 kg Final answer = 2.6 × 102 m / s. 5-17. The forces must have equal magnitudes, so Fastronaut = Fsatellite .

mastronaut aastronaut = msatellite asatellite . asatellite =
0.063 m/s . 5-18. t(s) 0 10 20 30 40
2

mastronaut aastronaut msatellite

=

95 kg × 0.50 m/s 2 = 750 kg Velocity (km/hr) 130 105 85 75 50...