a) A force F = (2.17 N/ m2 ) x2 i is applied to a particle initialy at rest in t

Question

a) A force F = (2.17 N/m2)x2i is applied to a particle initialy at rest in the xy plane. Calculate the work done by this force on the particle as it moves in a straight line from point (1.93 m, 1.93 m) to point (1.93 m, 6.89 m). The given force is the only force doing work on the particle. b) Calculate the work done on the particle as it moves along the path in a straight line from point (1.93 m, 1.93 m) to point (5.05 m, 5.93 m). For a), this is what I tried:
So (2.17 / 3) (6.89^3 - 1.93^3), (2.17 / 3) (327.08 - 7.189), (2.17 / 3) (319.89), = 231.39 W. So if W = F * d, W = 231.39 * (6.89 - 1.93) = 1147.7 J What am I doing wrong? The answer to a) is only necessary, as I can surely figure out b) if I figure out a). Thanks!

Explanation / Answer

Actually the given problem is theoritical rather than solving it by numerical method. Given force is F = (2.17 N/m^2) x^2 (i) i.e., the force is acting along x -axis. And as the particle moves from (1.93 m , 1.93 m ) to the point (1.93 m , 6.89 m ) by observing the co-ordinates of the particle , it is clear that the particel is movin gonly along y -axis as it 's y -co ordinate is changing. Hence both the applied force and the displace ment of the particle are perpendicular. Hence work done W = Fscos90 = Fs (0) = 0 Hence the work done is zero . No need of solving like that what you are doing . This the correct way to do the problem.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.