1. ΣFy = FT - Fg = ma FT = ma + mg = m (a + g) = 5 (0.3 + 9.8) FT
1. A 5.0-kg object is to be given an upward acceleration of
0.3 m/s2 by a rope pulling straight up on it. What is the tension
in the rope?
m = 5 kg
a = 0.3 m/s2
ΣFy = FT - Fg = ma
FT = ma + mg
= m (a + g)
= 5 ( )
= 50.5 N FT Fg

2. FN ΣFx = - Ff = ma - μ FN = ma - μ mg = ma a = - μ g = - 0.7(9.8)
2. A driver sees a horse on the road and applies the brakes so hard that they lock and the car skids to a stop in 24 m. The road is leveled, and the coefficient of kinetic friction between tires and road is 0.7. How fast was the car going when the brakes were applied? FN
ΣFx = - Ff = ma
- μ FN = ma
- μ mg = ma
a = - μ g = - 0.7(9.8)
= m/s2
x = 24 m
μk = 0.7 Ff Fg
= 18.1 m/s

3. FN FA Ff Fgy Fg ΣFy = FN - Fgy = 0 Fgx FN = Fgy Fgy = Fg cos θ
3. A 60 kg box slides down a plane 8 m long that is inclined at an angle of 30° with the horizontal. (μk = 0.25)
a. How much force is needed to move it up the plane at constant velocity? FN FA Ff θ
Fgy Fg
ΣFy = FN - Fgy = 0
FN = Fgy
Fgy = Fg cos θ
= 60(9.8) cos 30˚
= N
Fgx
m = 60 kg
θ = 30°
μk = 0.25

4. FN Fa ΣFx = Fa - Fgx - Ff = 0 Ff Fgy Fa = Fgx + Ff = Fg sin θ + μk FN
m = 60 kg
θ = 30°
μk = 0.25
FN = N

5. FN Fa Ff Fg Fgy ΣFx = Fa - Fgx - Ff = ma Fa = ma + Fgx + Ff
b. How much force is needed to move it up the plane with an acceleration of 2 m/s2?
a = 2 m/s2 Ff θ Fg
Fgy
ΣFx = Fa - Fgx - Ff = ma
Fa = ma + Fgx + Ff
= 60(2)
= N
Fgx

6. FN Ff ΣFx = Fgx - Ff = ma ΣFy = FN - Fgy = 0 Fgy Fg Fgx
4. A box slides down a plane 8 m long that is inclined at an angle of 30° with the horizontal. If the box starts from rest and µk = 0.25, find the acceleration of the box.
x = 8 m
θ = 30
vo = 0 m/s
μk = 0.25 FN Ff
ΣFx = Fgx - Ff = ma
ΣFy = FN - Fgy = 0 θ
Fgy Fg
Fgx

7. Substituting Fgx and Ff in ΣFx Ff = μkFN and FN = Fgy
Fgx = Fg sin θ
Fgy = Fg cos θ
ΣFx = Fgx - Ff = ma
Substituting Fgx and Ff in ΣFx
Ff = μkFN and FN = Fgy
Fg sin θ - μk Fg cos θ = ma
mg sin 30˚- μk mg cos 30˚ = ma
a = g (sin 30˚- μk cos 30˚)
a = 9.8 (sin 30˚- (0.25) cos 30˚)
= 2.8 m/s2 FN Ff θ
Fgy Fg
Fgx

8. 2000 B2. Blocks 1 and 2 of masses ml and m2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle  with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

9. a. On the figure below, draw and label all the forces on block ml.FN FT Ff Fg

10. Express your answers to each of the following in terms of ml, m2, g, , and f.
b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

11. c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane.
FN2 FT 2f
FN1 FT FT f
m2 g
m1 g Mg

12. d. The string between blocks 1 and 2 is now cut
d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.
FN2 FT 2f
FN1 FT FT f
m2 g
m1 g Mg

13. 5. An object mA = 25 kg rests on a tabletop
5. An object mA = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass
mB = 15 kg. If the coefficient of friction μk is 0.20 between the table and block 1, how far will block 2 drop in the first 3.0 s after the system is released?
m1 = 25 kg
m2= 15 kg
μk = 0.2
t = 3 s

14. FN = FgA = 25(9.8) = 245 N FfA = μFN = 0.2 (245) = 49 N m1 = 25 kg
μ = 0.2
t = 3 s FN
FfA FT FT
FN = FgA
= 25(9.8)
= 245 N
FfA = μFN
= 0.2 (245)
= 49 N
FgA
FgB

15. ΣF = FT - FfA + FgB - FT = mT aFN
FfA= 49 N FT
FfA FT
FgA
ΣF = FT - FfA + FgB - FT = mT a
FgB
= 2.45 m/s2
x = ½ at2
= ½ (2.45)(3)2
= 11 m

16. 2003 B1 A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on the floor. The opposite end of the rope is held by Student B of mass 60 kg, who is suspended at rest above the floor.
a. On the dots below that represent the students, draw and label free‑body diagrams showing the forces on Student A and on Student B. FT FT FN
FGA
FGB

17. b. Calculate the magnitude of the force exerted by the floor on Student A.FN
FGB
FGA
= 98 N

18. c. Student B now climbs up the rope at a constant acceleration of 0
c. Student B now climbs up the rope at a constant acceleration of 0.25 m/s2 with respect to the floor.
Calculate the tension in the rope while Student B is accelerating. FT FT FN
FGB
FGA
= 603 N
d. As Student B is accelerating, is Student A pulled upward off the floor? Justify your answer.
No. The minimum tension should be Student A’s weight: 686 N

19. FT
e. With what minimum acceleration must Student B climb up the rope to lift Student A upward off the floor? FT FN
FGB
FGA
= 1.63 m/s2

20. 1979B2. A 10‑kg block rests initially on a table as shown in cases I and II above. The coefficient of sliding friction between the block and the table is The block is connected to a cord of negligible mass, which hangs over a massless frictionless pulley. In case I a force of 50 N is applied to the cord. In case II an object of mass 5 kg is hung on the bottom of the cord. Use g = 10 m/s2.

21. a. Calculate the acceleration of the 10‑kg block in case I.

22. b. On the diagrams below, draw and label all the forces acting on each block in case II

23. c. Calculate the acceleration of the 10‑kg block in case II.

24. 1986B1. Three blocks of masses 1. 0, 2. 0, and 4
1986B1. Three blocks of masses 1.0, 2.0, and 4.0 kgs are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.
Calculate each of the following.
a. The acceleration of the 4‑kg block

25. b. The tension in the string supporting the 4‑kilogram block
c. The tension in the string connected to the l‑kilogram block

26. 1988B1. A helicopter holding a 70‑kg package suspended from a rope 5
1988B1. A helicopter holding a 70‑kg package suspended from a rope 5.0 m long accelerates upward at a rate of 5.2 m/s2. Neglect air resistance on the package.
a.On the diagram below, draw and label all of the forces acting on the package.

27. b. Determine the tension in the rope.

28. c. When the upward velocity of the helicopter is 30 m/s, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2.0 s after the rope is cut.