AP Physics 1 Unit 2 Forces MCQ Set B (Problem Solving)AP Physics 1 Unit 2 Forces MCQ Set B (Problem Solving)

APPHYSICS1 · U N I T 2 F O R C E S T E S T · MCQSETB

Solving MCQ

16 questions · calculator allowed · about 35 minutes · use g = 9.8 m/s². These are calculation questions on F = ma from graphs, elevators, friction, modified Atwood, angled pulls, inclines, springs, center of mass and connected objects. Draw the FBD and write ΣF = ma before you pick an answer. Each wrong choice is what one specific mistake gives you. The worked solutions are on a new page.

1. 2.5 A student pushes a 5.0 kg box horizontally across a floor. Friction on the box is 8.0 N. What push makes the box accelerate at 2.4 m/s²?

(A) 4.0 N (B) 12 N

(C) 20 N (D) 32 N

2. 2.5 · GRAPH

2468 10

10

20

30

t (s)

v (m/s)

(8, 24)(8, 24)

The graph shows the velocity of a 1200 kg car. During the first 8 s, air resistance on the car is 900 N. What forward force does the road exert on the car during that time?

(A) 2700 N (B) 3600 N

(C) 900 N (D) 4500 N

3. 2.6 · ELEVATOR

2 m/s²2 m/s²

A 60 kg student stands on a scale in an elevator that accelerates upward at 2.0 m/s². What does the scale read?

(A) 468 N (B) 588 N

(C) 708 N (D) 720 N

4. 2.7 · FRICTION A 15 kg crate is pulled horizontally with 80 N across a floor with μk = 0.30. What is its acceleration?

(A) 2.39 m/s² (B) 5.03 m/s²

(C) 5.33 m/s² (D) 8.27 m/s²

5. 2.7 · FRICTION A horizontal 60 N force accelerates an 8.0 kg block across a level surface at 4.5 m/s². What is the coefficient of kinetic friction?

(A) 0.40 (B) 0.31

(C) 0.46 (D) 0.77

6. 2.5 · MODIFIED ATWOOD

12 kg

4 kg

frictionlessfrictionless

A 12 kg block on a frictionless table is connected over an ideal pulley to a hanging 4.0 kg block. What is the tension in the string?

(A) 9.8 N (B) 29.4 N

(C) 39.2 N (D) 49.0 N

7. 2.7 · MODIFIED ATWOOD + FRICTION

10 kg

5 kg

μₖ = 0.20μₖ = 0.20

The same kind of setup now uses a 10 kg table block, a 5.0 kg hanging block, and μk = 0.20 between the table block and the table. What is the acceleration?

(A) 3.27 m/s² (B) 1.96 m/s²

(C) 2.61 m/s² (D) 4.90 m/s²

8. 2.7 · INCLINE

m

25°25°

vv

μₖ = 0.20μₖ = 0.20

A block slides down a 25° incline with μk = 0.20. What is its acceleration?

(A) 4.14 m/s² (B) 5.92 m/s²

(C) 2.37 m/s² (D) 1.78 m/s²

9. 2.4 · ANGLED PULL

20 kg

50 N50 N

30°30°

frictionlessfrictionless

A 20 kg box is pulled across a frictionless floor by a 50 N force at 30° above horizontal. What is the normal force on the box?

(A) 196 N (B) 221 N

(C) 153 N (D) 171 N

10. 2.7 · ANGLED PULL + FRICTION

15 kg

60 N60 N

37°37°

μₖ = 0.25μₖ = 0.25

A 15 kg box is pulled by a 60 N force at 37° above horizontal across a floor with μk = 0.25. What is its acceleration?

(A) 1.35 m/s² (B) 0.74 m/s²

(C) 3.19 m/s² (D) 0.14 m/s²

11. 2.8 · SPRINGS A 2.0 kg mass hangs at rest from a spring, stretching it 0.15 m. What is the spring constant? (A) 13.3 N/m (B) 2.94 N/m

(C) 131 N/m (D) 196 N/m

12. 2.8 · SPRING GRAPH

0.1 0.2 0.3 0.4

4

8

12

16

20

x (m)

F (N)

(0.10, 5)(0.10, 5)

(0.30, 15)(0.30, 15)

The graph shows the force needed to stretch a spring. What mass, hung at rest from this spring, would stretch it 0.30 m?

(A) 0.153 kg (B) 1.53 kg

(C) 15.0 kg (D) 150 kg

13. 2.1 · CENTER OF MASS

−112345

−1

1

2

3

4

x (m)

y (m)

2 kg2 kg 3 kg3 kg

5 kg5 kg

Three small masses are placed as shown. Where is the system's center of mass?

(A) (2.0, 1.0) (B) (2.2, 1.5)

(C) (2.0, 1.5) (D) (2.2, 1.0)

14. 2.3 · CONNECTED OBJECTS

6 kg 4 kg

30 N30 N

frictionless icefrictionless ice

A horizontal 30 N force pushes a 6.0 kg block, which pushes a 4.0 kg block ahead of it across frictionless ice. What is the magnitude of the force the 6.0 kg block exerts on the 4.0 kg block?

(A) 12 N (B) 18 N

(C) 15 N (D) 30 N

15. 2.6 · ELEVATOR, TWO BLOCKS

T₁ = 1200 NT₁ = 1200 N

30 kg

50 kg

T₁T₁

T₂T₂ aa

Two blocks hang from the ceiling of an accelerating elevator: a 30 kg block with a 50 kg block hanging from it. The tension in the top cable is 1200 N. What is the tension T₂ in the lower cable?

(A) 490 N (B) 450 N

(C) 1200 N (D) 750 N

16. 1.X + 2.7A2.0 kg puck slides across a floor at 6.0 m/s and comes to a stop in 3.0 m. What is the coefficient of kinetic friction?

(A) 0.61 (B) 0.20

(C) 0.33 (D) 1.0

Answer key with worked solutions

123456789 10 11 12 13 14 15 16

C D C A B B B C D A C B B A D A

1. (C) FA − f = ma, so FA = 5(2.4) + 8 = 12 + 8 = 20 N. (B) leaves out friction. (A) subtracts friction instead of adding it. 2. (D) a = 24/8 = 3 m/s². Then F − Fdrag = ma, so F = 1200(3) + 900 = 4500 N. (B) is only the net force. (A) subtracts the drag.

3. (C) Up is +. FN − mg = ma, so FN = 60(9.8 + 2.0) = 708 N. (A) uses a downward acceleration. (B) is mg. (D) uses g = 10.

4. (A) f = 0.30(15)(9.8) = 44.1 N, so a = (80 − 44.1)/15 = 2.39 m/s². (B) uses μm instead of μmg. (C) leaves out friction. (D) adds friction.

5. (B) 60 − f = 8(4.5) = 36, so f = 24 N. Then μk = 24/(8 × 9.8) = 0.31. (A) divides by the applied force. (D) uses 60/FN.

6. (B) a = mg/(m + M) = 4(9.8)/16 = 2.45 m/s². Then FT = Ma = 12(2.45) = 29.4 N. Check: m(g − a) = 4(7.35) = 29.4 ✓. (C) is mg, but the tension has to be less than that.

7. (B) a = g(m − μM)/(m + M) = 9.8(5 − 0.2·10)/15 = 9.8(3)/15 = 1.96 m/s². (A) leaves out friction. (D) divides by the hanging mass only.

8. (C) a = g(sin25° − 0.2 cos25°) = 9.8(0.4226 − 0.1813) = 2.37 m/s². (A) leaves out friction. (B) adds friction, which would be right for a block sliding up.

9. (D) FN + 50 sin30° − 196 = 0, so FN = 196 − 25 = 171 N. (A) assumes FN = mg. (B) treats it as a push down. (C) uses cos instead of sin.

10. (A) Fx = 60cos37° = 47.9 N and Fy = 36.1 N. Then FN = 147 − 36.1 = 110.9 N and f = 0.25(110.9) = 27.7 N, so a = (47.9 − 27.7)/15 = 1.35 m/s². (B) uses FN = mg. (C) leaves out friction. (D) treats it as a push down. 11. (C) kx = mg, so k = 19.6/0.15 = 131 N/m. (A) uses mass instead of weight. (B) multiplies instead of dividing. 12. (B) k = slope = (15 − 5)/(0.30 − 0.10) = 50 N/m. At 0.30 m, F = 15 N = mg, so m = 15/9.8 = 1.53 kg. (C) treats the force as the mass.