Chapter 7. Newton's Second Law

Chapter 7. Newton’s Second Law

Newton’s Second Law

7-1. A 4-kg mass is acted on by a resultant force of (a) 4 N, (b) 8 N, and (c) 12 N. What are the resulting accelerations?

(a) 1 m/s2 (b) 2 m/s2 (c) 3 m/s2

7-2. A constant force of 20 N acts on a mass of (a) 2 kg, (b) 4 kg, and (c) 6 kg. What are the resulting accelerations?

(a) 10 m/s2 (b) 5 m/s2 (c) 3.33 m/s2

7-3. A constant force of 60 lb acts on each of three objects, producing accelerations of 4, 8, and 12 N. What are the masses?

15 slugs 7.5 slugs 5 slugs

7-4. What resultant force is necessary to give a 4-kg hammer an acceleration of 6 m/s2?

F = ma = (4 kg)(6 m/s2); F = 24 N

7-5. It is determined that a resultant force of 60 N will give a wagon an acceleration of 10 m/s2. What force is required to give the wagon an acceleration of only 2 m/s2?

; F = ma = (6 slugs)(2 m/s2); F = 12 N

7-6. A 1000-kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude and direction of the force? Convert to SI units: 100 km/h = 27.8 m/s

F = ma = (1000 kg)(7.72 m/s2); F = 772 N, South.

The Relationship Between Weight and Mass

7-7. What is the weight of a 4.8 kg mailbox? What is the mass of a 40-N tank?

W = (4.8 kg)(9.8 m/s2) = 47.0 N ; = 4.08 kg

7-8. What is the mass of a 60-lb child? What is the weight of a 7-slug man?

= 1.88 slugs ; W = (7 slugs)(32 ft/s2) = 224 lb

7-9. A woman weighs 180 lb on earth. When she walks on the moon, she weighs only 30 lb. What is the acceleration due to gravity on the moon and what is her mass on the moon? On the Earth?

Her mass is the same on the moon as it is on the earth, so we first find the constant mass:

mm = me = 5.62 slugs ;

Wm = mmgm ; gm = 5.33 ft/s2

7-10. What is the weight of a 70-kg astronaut on the surface of the earth. Compare the resultant force required to give him or her an acceleration of 4 m/s2 on the earth with the resultant force required to give the same acceleration in space where gravity is negligible?

On earth: W = (70 kg)(9.8 m/s2) = 686 N ; FR = (70 kg)(4 m/s2) = 280 N

Anywhere: FR = 280 N The mass doesn’t change.

7-11. Find the mass and the weight of a body if a resultant force of 16 N will give it an acceleration of 5 m/s2.

= 3.20 kg ; W = (3.20 kg)(9.8 m/s2) = 31.4 N

7-12. Find the mass and weight of a body if a resultant force of 200 lb causes its speed to increase from 20 ft/s to 60 ft/s in a time of 5 s.

= 25.0 slugs

W = mg = (25.0 slugs)(32 ft/s2); W = 800 lb

7-13. Find the mass and weight of a body if a resultant force of 400 N causes it to decrease its velocity by 4 m/s in 3 s.

; m = 300 kg

W = mg = (300 kg)(9.8 m/s2); W = 2940 N

Applications for Single-Body Problems:

7-14. What horizontal pull is required to drag a 6-kg sled with an acceleration of 4 m/s2 if a friction force of 20 N opposes the motion?

P – 20 N = (6 kg)(4 m/s2); P = 44.0 N

7-15. A 2500-lb automobile is speeding at 55 mi/h. What resultant force is required to stop the car in 200 ft on a level road. What must be the coefficient of kinetic friction?

We first find the mass and then the acceleration. (55 mi/h = 80.7 m/s)

F = ma = (78.1 slugs)(-16.3 ft/s2); F = -1270 lb

k = 0.508

7-16. A 10-kg mass is lifted upward by a light cable. What is the tension in the cable if the acceleration is (a) zero, (b) 6 m/s2 upward, and (c) 6 m/s2 downward?

Note that up is positive and that W = (10 kg)(9.8 m/s2) = 98 N.

(a) T – 98 N = (10 kg)(0 m/sand T = 98 N

(b) T – 98 N = (10 kg)(6 m/sand T = 60 N + 98 N or T = 158 N

(c) T – 98 N = (10 kg)(-6 m/sand T = - 60 N + 98 N or T = 38.0 N

7-17. A 64-lb load hangs at the end of a rope. Find the acceleration of the load if the tension in the cable is (a) 64 lb, (b) 40 lb, and (c) 96 lb.

(a) ; a = 0

(b) ; a = -12.0 ft/s2

(b) ; a = 16.0 ft/s2

7-18. An 800-kg elevator is lifted vertically by a strong rope. Find the acceleration of the elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N.

Newton’s law for the problem is: T – mg = ma (up is positive)

(a) 9000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = 1.45 m/s2

(a) 7840 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = 0

(a) 2000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = -7.30 m/s2

7-19. A horizontal force of 100 N pulls an 8-kg cabinet across a level floor. Find the acceleration of the cabinet if k = 0.2.

F = kN = k mg F = 0.2(8 kg)(9.8 m/s

100 N – F = ma; 100 N – 15.7 N = (8 kg) a; a = 10.5 m/s2

7-20. In Fig. 7-10, an unknown mass slides down the 300 inclined plane.

What is the acceleration in the absence of friction?

Fx = max; mg sin 300 = ma ; a = g sin 300

a = (9.8 m/s2) sin 300 = 4.90 m/s2, down the plane

7-21. Assume that k = 0.2 in Fig 7-10. What is the acceleration?

Why did you not need to know the mass of the block?

Fx = max; mg sin 300 - kN = ma ; N = mg cos 300

mg sin 300 - k mg cos 300 = ma ; a = g sin 300 - k g cos 300

a = (9.8 m/s2)(0.5) – 0.2(9.8 m/s2)(0.866); a = 3.20 m/s2, down the plane.

*7-22. Assume that m = I 0 kg and k = 0. 3 in Fig. 7- 10. What push P directed up and along the incline in Fig.7-10 will produce an acceleration of 4 m/s2 also up the incline?

F = kN = kmg cos 300; F = 0.3(10 kg)(9.8 m/s2)cos 300 = 25.5 N

Fx = ma; P – F – mg sin 300 = ma

P – 25.5 N – (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2)

P – 25.5 N – 49.0 N = 40 N; P = 114 N

*7-23. What force P down the incline in Fig. 7-10 is required to cause the acceleration DOWN the plane to be 4 m/s2? Assume that in = IO kg and k = 0. 3.

See Prob. 7-22: F is up the plane now. P is down plane (+).

Fx = ma; P - F + mg sin 300 = ma ; Still, F = 25.5 N

P - 25.5 N + (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2)

P - 25.5 N + 49.0 N = 40

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