a) select m or m/s b) select +x or -x c) select m or m/s d) select +x or -c The

Question

a) select m or m/s b) select +x or -x c) select m or m/s d) select +x or -c The figure below shows a plot of potential energy U versus position x of a 0.94 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U-20.0 J, Ue-450 , and UC-55.0 . 6 x (m) The particle is released at x -4.5 m with an initial speed of 8.0 m/s, headed in the negative x direction. (a) If the particle can reach x-1.0 m, what is its speed there, and if it cannot, what is its turning point? Select… (b) what are the magnitude and direction of the force on the narticle as it begins to move to the left of x = 4.0 m? magnitude direction … select… Suppose, instead, the particle is headed in the positive x direction when it is released at x 4.5 m at speed 8.0 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? (d) what are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? magnitude direction Select

Explanation / Answer

Ua = 20J, Ub = 45J, Uc = 55J

a)Initial potential energy = 20J

Final potential energy = 45J

Kinetic energy = 0.5mv^2

=0.5*0.94*8^2

=30.08 J

since Ua+Initial kinetic energy > Ub, it will reach there.

so the kinetic energy = 20+30-45

=5 Joules

so, 0.5mv^2=5

or 0.5*0.94*v^2 = 5

or v=3.26 m/s

b)Force = (-dU/dx)

= -slope.

slope = (20-45)/(4-2)

=12.5 N (towards positive x axis)

c)since Ua+initial kinetic energy < Uc, it will not reach the point 7m.

force acting on the particle = (-dU/dx)

=-slope

=-(55-20)/(6-5)

=-35 N

so force = 35N towards negative x axis.

let the distance stopped be x.so,

KEini = Work done by the force

or 30 = 35*(x)

or x = 0.857

so distance on the x-axis of its return point = 5+0.857

=5.857 m

d)Force = 35 N (in the negative x axis)

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