1. Work
A constant force F = 10 N acts on a particle that moves a distance d = 3 m, with angle θ between the force and the displacement. Which expression gives the work done by the force?
| A. W = Fd cosθ | B. W = Fd sinθ | C. W = Fd |
| D. W = F/d | E. W = F cosθ/d | F. W = d cosθ/F |
2. Work
A force F = 8 N acts perpendicular to the displacement of a particle of length d = 2 m. Which expression gives the work done by the force?
| A. W = Fd | B. W = Fd sinθ | C. W = Fd cosθ |
| D. W = 0 | E. W = F/d | F. W = d/F |
3. Work done by a variable force
A particle moves along the x-axis from x₀ = 1 m to x = 4 m under a variable force F(x). Which expression gives the work done by the force?
A. W = ∫ₓ₀ˣ F(x) dx
B. W = F(x)(x − x₀)
C. W = dF/dx
D. W = ∫ₓ₀ˣ x dF
E. W = F(x)/x
F. W = ∫ₓ₀ˣ F(t) dt
4. Work done by a variable force
A force depends on position as F(x) = (3 N/m)x. The particle moves from x = 0 m to x = 2 m. Which expression gives the work?
| A. W = ∫₀² 3x dx | B. W = 3(2 − 0) |
| C. W = d(3x)/dx | D. W = ∫₀² x d(3x) |
| E. W = 3x² | F. W = 3/x |
5. Net work and kinetic energy
A particle of mass m changes speed from v₀ = 2 m/s to v = 5 m/s. Which expression gives the net work done on the particle?
| A. W = K + K₀ | B. W = K − K₀ | C. W = K₀ − K |
| D. W = mv − mv₀ | E. W = ½m(v − v₀)² | F. W = ½m(v + v₀)² |
6. Net work and kinetic energy
A particle of mass m speeds up from v₀ = 2 m/s to v = 4 m/s. Which expression gives the net work done on it?
| A. W = ½m(4² − 2²) | B. W = ½m(4 − 2)² |
| C. W = m(4 − 2) | D. W = ½m(4² + 2²) |
| E. W = ½m(2² − 4²) | F. W = m(4 + 2) |
7. Work-energy theorem on an incline with friction
A block of mass m = 2 kg moves up an incline of angle θ through distance s = 5 m. The kinetic friction coefficient is μₖ = 0.20. What is the net work done on the block by gravity and friction during the upward motion?
A. W = mg s sinθ + μₖmg s cosθ
B. W = −mg s sinθ + μₖmg s cosθ
C. W = −mg s sinθ − μₖmg s cosθ
D. W = mg s cosθ − μₖmg s sinθ
E. W = −mg s cosθ − μₖmg s sinθ
F. W = μₖmg s
8. Work-energy theorem on an incline with friction
A block of mass m = 3 kg slides down an incline of angle θ through distance s = 4 m. The kinetic friction coefficient is μₖ = 0.10. Which expression gives the net work done by gravity and friction?
A. W = −mg s sinθ − μₖmg s cosθ
B. W = mg s sinθ − μₖmg s cosθ
C. W = mg s cosθ − μₖmg s sinθ
D. W = mg s sinθ + μₖmg s cosθ
E. W = −mg s sinθ + μₖmg s cosθ
F. W = μₖmg s
9. Work in quasistatic spring compression
A spring with constant k = 200 N/m is compressed quasistatically from x₀ = 0.10 m to x = 0.30 m, where x > x₀ > 0. Which expression gives the work done by the external agent?
| A. W = −½k(x² − x₀²) | B. W = ½k(x² − x₀²) | C. W = k(x − x₀) |
| D. W = ½k(x − x₀)² | E. W = k(x² − x₀²) | F. W = −k(x − x₀) |
10. Work in quasistatic spring compression
A spring with constant k = 150 N/m is compressed quasistatically from x₀ = 0.05 m to x = 0.20 m, where x > x₀ > 0. Which expression gives the work done by the spring?
| A. W = ½k(x² − x₀²) | B. W = −½k(x² − x₀²) | C. W = k(x − x₀) |
| D. W = −k(x − x₀) | E. W = ½k(x − x₀)² | F. W = k(x² + x₀²) |
11. Work as a line integral
A particle moves along a curved path of length 6 m in a force field F. Which expression gives the work done by the force along the path?
| A. W = ∫ F dr | B. W = ∫ F · dr | C. W = ∫ dr/F |
| D. W = F · r | E. W = dF/dr | F. W = ∫ r · dF |
12. Work as a line integral
A particle moves from point 1 to point 2 along a path C in a force field F(r). The displacement along the path is measured in meters. Which expression gives the work done?
A. W = ∫C F · dr
B. W = ∫C F dr
C. W = F · r
D. W = dF/dr
E. W = ∫C r · dF
F. W = ∫C dr/F
13. Power in lifting with tension
A box is lifted straight upward at speed v = 2 m/s by a rope with tension T = 50 N. Which expression gives the instantaneous power delivered by the rope?
| A. P = Tv | B. P = mgv | C. P = T/v |
| D. P = T + v | E. P = Tv cos90° | F. P = Tm |
14. Power in lifting with tension
A box is lifted vertically with tension T = 40 N and upward velocity v = 3 m/s. If the speed is constant, which expression gives the power delivered by the rope?
| A. P = Tv | B. P = mg/v | C. P = T/v |
| D. P = T + v | E. P = 0 | F. P = Tv² |
15. Power given as a function of time
The power delivered to a system varies with time as P(t). Time t is measured in seconds. Which expression gives the work done from t = 0 s to t?
| A. W = P(t)t | B. W = dP/dt | C. W = ∫₀ᵗ P(t) dt |
| D. W = P/t | E. W = ∫₀ᵗ t dP | F. W = Pt² |
16. Power given as a function of time
The instantaneous power is P(t) = (2 W/s)t. Which expression gives the work done from t = 0 s to t = 3 s?
| A. W = ∫₀³ 2t dt | B. W = 2(3) |
| C. W = d(2t)/dt | D. W = ∫₀³ t d(2t) |
| E. W = 2t² | F. W = 2/t |
17. Work done by friction
A block slides a distance d = 6 m on a horizontal surface with kinetic friction magnitude fₖ = 4 N. Which expression gives the work done by friction on the block?
| A. W = fₖd | B. W = −fₖd | C. W = μₖd |
| D. W = −μₖmg/d | E. W = fₖ/d | F. W = 0 |
18. Work done by friction
A block moves to the right on a horizontal surface. Kinetic friction of magnitude μₖmg acts on it, where μₖ = 0.30, m = 2 kg, and the distance is d = 5 m. Which expression gives the work done by friction over distance d?
A. W = μₖmgd
B. W = −μₖmgd
C. W = μₖgd
D. W = −μₖgd
E. W = μₖmg/d
F. W = 0
19. Power from P = Fv
A particle moves with velocity v = 4 m/s in the same direction as a constant force F = 12 N. Which expression gives the instantaneous mechanical power delivered by the force?
| A. P = F/v | B. P = Fv | C. P = F + v |
| D. P = Fv² | E. P = F²v | F. P = Fv cos90° |
20. Power from P = Fv
A constant force F = 9 N acts opposite to the velocity v = 3 m/s of a moving particle. Which expression gives the instantaneous power delivered by the force?
| A. P = Fv | B. P = −Fv | C. P = F/v |
| D. P = 0 | E. P = F + v | F. P = Fv² |
Correct answers:
1. A
2. D
3. A
4. A
5. B
6. A
7. C
8. B
9. B
10. B
11. B
12. A
13. A
14. A
15. C
16. A
17. B
18. B
19. B
20. B
