The figure here shows a plot of potential energy U versus position x of a 0.90 k

Question

The figure here shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x-axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the right of = 5.0 m? A block of mass m is dropped onto a spring with spring constant k from a height h. The block becomes attached to the spring and compresses it by distance d before momentarily stopping. Your answer must he in terms of nt, g, h, and k. Be sure to show all your work.

Explanation / Answer

1)a)Total energy at x=4.5=15J+0.5*0.9*7^2=37.05J

By conservation of energy, Kinetic energy at x=1=37.05-35J=2.05J

0.5*0.9*v^2=2.05

v=2.134m/s

b)F=-du/dx=(35-15)/2=10N

Direction is towards the +x axis.

c)At turning point, all initial energy=potential energy,

U=37.05J

U between 5 and 6 is given by

U=15+30*(x-5)=37.05

Solving for x,

x=5.735m

Therefore, the partical turns at x=5.735m

d)F=-dU/dx=(45-15)/1=-30N

Direction is to the left.

2)Using energy conservation,

mg(h+d)=0.5kd^2

d^2-2mg/k(h+d)=0

Solving for d,

d=sqrt((2mg/k)^2+8mgh/k)/2

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