Work-Energy Problems

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