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06: Applications of Newton's Laws

Check Your Understanding

6.1. Fs = 645 N

6.2. a = 3.68 m/s2, T = 18.4 N

6.3. T = \(\frac{2m_{1}m_{2}}{m_{1} + m_{2}}\)g (This is found by substituting the equation for acceleration into the equation for tension in Figure 6.7.)

6.4. 1.49 s

6.5. 49.4 degrees

6.6. 128 m; no

6.7. a. 4.9 N; b. 0.98 m/s2

6.8. −0.23 m/s2; the negative sign indicates that the snowboarder is slowing down.

6.9. 0.40

6.10. 34 m/s

6.11. 0.27 kg/m

Conceptual Questions

1. The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

3. If you do not let up on the brake pedal, the car’s wheels will lock so that they are not rolling; sliding friction is now involved and the sudden change (due to the larger force of static friction) causes the jerk.

5. 5.00 N

7. Centripetal force is defined as any net force causing uniform circular motion. The centripetal force is not a new kind of force. The label “centripetal” refers to any force that keeps something turning in a circle. That force could be tension, gravity, friction, electrical attraction, the normal force, or any other force. Any combination of these could be the source of centripetal force, for example, the centripetal force at the top of the path of a tetherball swung through a vertical circle is the result of both tension and gravity.

9. The driver who cuts the corner (on Path 2) has a more gradual curve, with a larger radius. That one will be the better racing line. If the driver goes too fast around a corner using a racing line, he will still slide off the track; the key is to stay at the maximum value of static friction. So, the driver wants maximum possible speed and maximum friction. Consider the equation for centripetal force: Fc = m\(\frac{v^{2}}{r}\) where v is speed and r is the radius of curvature. So by decreasing the curvature (\(\frac{1}{r}\)) of the path that the car takes, we reduce the amount of force the tires have to exert on the road, meaning we can now increase the speed, v. Looking at this from the point of view of the driver on Path 1, we can reason this way: the sharper the turn, the smaller the turning circle; the smaller the turning circle, the larger is the required centripetal force. If this centripetal force is not exerted, the result is a skid.

11. The barrel of the dryer provides a centripetal force on the clothes (including the water droplets) to keep them moving in a circular path. As a water droplet comes to one of the holes in the barrel, it will move in a path tangent to the circle.

13. If there is no friction, then there is no centripetal force. This means that the lunch box will move along a path tangent to the circle, and thus follows path B. The dust trail will be straight. This is a result of Newton’s first law of motion.

15. There must be a centripetal force to maintain the circular motion; this is provided by the nail at the center. Newton’s third law explains the phenomenon. The action force is the force of the string on the mass; the reaction force is the force of the mass on the string. This reaction force causes the string to stretch.

17. Since the radial friction with the tires supplies the centripetal force, and friction is nearly 0 when the car encounters the ice, the car will obey Newton’s first law and go off the road in a straight line path, tangent to the curve. A common misconception is that the car will follow a curved path off the road.

19. Anna is correct. The satellite is freely falling toward Earth due to gravity, even though gravity is weaker at the altitude of the satellite, and g is not 9.80 m/s2. Free fall does not depend on the value of g; that is, you could experience free fall on Mars if you jumped off Olympus Mons (the tallest volcano in the solar system).

21. The pros of wearing body suits include: (1) the body suit reduces the drag force on the swimmer and the athlete can move more easily; (2) the tightness of the suit reduces the surface area of the athlete, and even though this is a small amount, it can make a difference in performance time. The cons of wearing body suits are: (1) The tightness of the suits can induce cramping and breathing problems. (2) Heat will be retained and thus the athlete could overheat during a long period of use.

23. The oil is less dense than the water and so rises to the top when a light rain falls and collects on the road. This creates a dangerous situation in which friction is greatly lowered, and so a car can lose control. In a heavy rain, the oil is dispersed and does not affect the motion of cars as much.


25. a. 170 N

b. 170 N

27. \(\vec{F}_{3} = (− 7\; \hat{i} + 2\; \hat{j} + 4\; \hat{k})\; N\)

29. 376 N pointing up (along the dashed line in the figure); the force is used to raise the heel of the foot.

31. −68.5 N

33. a. 7.70 m/s2 ; b. 4.33 s

35. a. 46.4 m/s

b. 2.40 x 103 m/s2

c. 5.99 x 103 N; ratio of 245

37. a. 1.87 x 104 N

b. 1.67 x 104 N

c. 1.56 x 104 N

d. 19.4 m, 0 m/s

39. a. 10 kg

b. 90 N

c. 98 N

d. 0

41. a. 3.35 m/s2

b. 4.2 s

43. a. 2.0 m/s2

b. 7.8 N

c. 2.0 m/s

45. a. 0.933 m/s2 (mass 1 accelerates up the ramp as mass 2 falls with the same acceleration)

b. 21.5 N

47. a. 10.0 N

b. 97.0 N

49. a. 4.9 m/s2

b. The cabinet will not slip.

c. The cabinet will slip.

51. a. 32.3 N, 35.2°

b. 0

c. 0.301 m/s2 in the direction of \(\vec{F}_{tot}\)

53. $$\begin{split} net\; F_{y} & = 0 \Rightarrow N = mg \cos \theta \\ net\; F_{x} & = ma \\ a & = g(\sin \theta − \mu k \cos \theta) \end{split}$$

55. a. 1.69 m/s2

b. 5.71°

57. a. 10.8 m/s2

b. 7.85 m/s2

c. 2.00 m/s2

59. a. 9.09 m/s2

b. 6.16 m/s2

c. 0.294 m/s2

61. a. 272 N, 512 N

b. 0.268

63. a. 46.5 N

b. 0.629 m/s2

65. a. 483 N

b. 17.4 N

c. 2.24, 0.0807

67. 4.14°

69. a. 24.6 m

b. 36.6 m/s2

c. 3.73 times g

71. a. 16.2 m/s

b. 0.234

73. a. 179 N

b. 290 N

c. 8.3 m/s

75. 20.7 m/s

77. 21 m/s

79. 115 m/s or 414 km/h

81. vT = 25 m/s; v2 = 9.9 m/s

83. \(\left(\dfrac{110}{65}\right)^{2}\) = 2.86 times

85. Stokes’ law is Fs = 6\(\pi\)r\(\eta\)v. Solving for the viscosity, \(\eta = \frac{F_{s}}{6 \pi rv}\). Considering only the units, this becomes [\(\eta\)] = \(\frac{kg}{m \cdotp s}\).

           87. 0.76 kg/m • s

           89. a. 0.049 kg/s

b. 0.57 m

91. a. 1860 N, 2.53

b. The value (1860 N) is more force than you expect to experience on an elevator. The force of 1860 N is 418 pounds, compared to the force on a typical elevator of 904 N (which is about 203 pounds); this is calculated for a speed from 0 to 10 miles per hour, which is about 4.5 m/s, in 2.00 s).

c. The acceleration a = 1.53 x g is much higher than any standard elevator. The final speed is too large (30.0 m/s is VERY fast)! The time of 2.00 s is not unreasonable for an elevator.

93. 189 N

95. 15 N

97. 12 N

Additional Problems

99. ax = 0.40 m/s2 and T = 11.2 x 103 N

101. m(6pt + 2q)

103. \(\vec{v}\)(t) = \(\left(\dfrac{pt}{m} + \dfrac{nt^{2}}{2m}\right) \hat{i} + \left(\dfrac{qt^{2}}{2}\right) \hat{j}\) and \(\vec{r}\)(t) = \(\left(\dfrac{pt^{2}}{2m} + \dfrac{nt^{3}}{6m}\right) \hat{i} + \left(\dfrac{qt^{3}}{60\; m}\right) \hat{j}\) 

105. 9.2 m/s

107. 1.3 s

109. 5.4 m/s2

111. a. 0.60

b. 1200 N

c. 1.2 m/s2 and 1080 N

d. −1.2 m/s2

e. 120 N

113. 0.789

115. a. 0.186 N

b. 774 N

c. 0.48 N

117. 13 m/s

119. 20.7 m/s

121. a. 28,300 N

b. 2540 m

123. 25 N

125. a = \(\frac{F}{4}\) − \(\mu_{k}\)g

127. 14 m

Challenge Problems

129. v = \(\sqrt{v_{0}^{2} − 2gr_{0} \left(1 − \dfrac{r_{0}}{r}\right)}\) 

131. 78.7 m

133. a. 53.9 m/s

b. 328 m

c. 4.58 m/s

d. 257 s

135. a. v = 20.0(1 − e−0.01t)

b. vlimiting = 20 m/s


  • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).