A particle is uncharged and is thrown vertically upward from groundlevel with a

Question

A particle is uncharged and is thrown vertically upward from groundlevel with a speed of 20.0 m/s. As a result, it attains a maximumheight h. The particle is then given a positive charge+q and reaches the same maximum height h whenthrown vertically upward with a speed of 25.0 m/s. The electric potential at the heighth exceeds the electric potential at ground level. Finally,the particle is given a negative charge -q. Ignoring airresistance, determine the speed with which the negatively chargedparticle must be thrown vertically upward, so that it attainsexactly the maximum height h. In all three situations, besure to include the effect of gravity.

Explanation / Answer

cute question...   call the threespeeds    u v w   so that u=20 v=25  w is unknown. . Then write conservation of energy for the threesituations: . initial KE = final grav PE + final electrical PE . first situation:    (1/2) mu2   =    m gh    + 0 . second situation:   (1/2) m v2 = m g h   + q V . third situation:    (1/2) m w2 = m g h   - qV . Multiply all three by 2 and divide by m: .            u2 = 2gh   .            v2 = 2gh + 2qV/m .           w2   = 2gh  - 2qV/m . Eliminate 2gh by substitution in the latter twoequations: .          v2 = u2   + 2qV/m .          w2 = u2 - 2qV/m . Add these two equations... note thatthe   2qV/m terms disappear: .     v2 +  w2   = 2u2 .     w = [ 2u2 - v2 ]1/2 = [ 2 *202 - 252]1/2   =    13.2 m/s    is the third speed

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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