6.7. Newton’s Laws Problem Set http://www.ck12.org
As you know, if the situation is left like this, the little block will just fall. But if we accelerate the leftmost
block to the right, this will create a normal force between the little block and the big block, and if there is a
coefficient of friction between them, then the little block won’t slide down! Clever, eh?
a. The mass of the little block is 0.15 kg. What frictional force is required to keep it from falling? (State a
magnitude and direction.)
b. If both blocks are accelerating to the right with an accelerationa= 14 .0 m/s^2 , what is the normal force
on the little block provided by the big block?
c. What is the minimum coefficient of static friction required?Answers to Selected Problems
1..
2..
3..
4. Zero; weight of the hammer minus the air resistance.
5. 2 forces
6. 1 force
7. No
8. The towel’s inertia resists the acceleration
9. a. Same distance b. You go farther c. Same amount of force
10..
11. a. 98 N b. 98 N
12..
13. 32 N
14. 5.7 m/s^2
15..
16..
17.Fx=14 N,Fy=20 N
18. Left picture:F=23N 98◦, right picture:F=54 N 5◦
19. 3 m/s^2 east
20. 4 m/s^2 ; 22. 5 ◦NE
21. 0. 51
22. 0. 2
23. The rope will not break because his weight of 784 N is distributed between the two ropes.
24. Yes, because his weight of 784 N is greater than what the rope can hold.
25. Mass is 51 kg and weight is 82 N
26. a. While accelerating down b. 686 N c. 826 N
27. a. 390 N b. 490 N
28. 0. 33
29. 3.6 kg
30. g sinθ
31. b.20 N c. 4.9 N d. 1.63 kg e. Eraser would slip down the wall
32. a. 1450 N b. 5600 N c. 5700 N d. Friction between the tires and the ground e. Fuel, engine, or equal and
opposite reaction
33. b. 210 N c. no, the box is flat so the normal force doesn’t change d. 2.8 m/s^2 e.28 m/s f. no g. 69 N h. 57 N
i. 40 N j. 0.33 k. 0. 09
34..
35. a. zero b.−kx 0
36. b.f 1 =μkm 1 g cosθ;f 2 =μkm 2 g cosθc. Ma d.TA= (m 1 +m 2 )(a+μcosθ)andTB=m 2 a+μm 2 cosθe. Solve