Problem 2 A particle is in a region where the potential energy has the form U =

Question

Problem 2 A particle is in a region where the potential energy has the form U = 9:0 x Jm: (Hint: the unit of meters is there to cancel out with the unit on x.) The particle starts out at rest at x = 0:45m:

(a)What is the particle's initial kinetic energy?

(b)Which way will it go if it is released from rest in the region where this equation describes the potential energy, and why?

(c)What is the particle's kinetic energy, if it is released from rest, when it passes through a point 1:35m from its starting point?

Explanation / Answer

given
potential eneregy is of the form U(x) = 9x

a. the particle starts at rest from x = 0.45 m
   so initial KE = 0
   as the particle starts from rest

b. PE at x = 0.45 m
   U(0.45) = 9*0.45 = 4.05 J
   so the particle would move to decrease the potnetial energy
   hence the particle would move in -ve x direciton reducing the value of x to decrease its potential energy

c. the particle will never go to x = 0.45 + 1.35 = 1.8 m if it was at rest initially at 0.45 m
   but the particle can move to x = 0.45 - 1.35 = -0.9 m
   PE at x = -0.9 m
   U(-0.9) = -9*0.9 = -8.1 J
   KE = ?
   so KE + PE = initial PE + iniitial KE
   KE = 8.1 + 0.45 = 8.55 J

   so KE of particle at x = -0.9 m ( 1.35 m away from the initial point) is 8.55 J

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