Question

The figure below shows a plot of potential energy U versus position x of a 1.04 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U₁ = 15 J, U₂ = 30 J, and U₃ = 40 J.

The figure below shows a plot of potential energy U versus position x of a 1.04 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 15), U) = 30 ), and U, = 40 ). U (J) - - 2 4 6 x (m) The particle is released at x = 4.5 m with an initial speed of 9.0 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? mis (b) 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? N in the positive direction. Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4,5 m at speed 9.0 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? mis (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? x N in the negative direction. The mechanical energy (the sum of K and U) remains constant as the particle moves. The graph gives U. You can thus calculate K, which cannot be negative. From K, you can calculate speed. A turning point is reached if K goes to zero. There a particle stops and reverses its motion. The force slowing, stopping, and reversing the particle is equal to the negative of the graph's slope.

Answer 1

a] Total mechanical energy is conserved,

Final PE + KE = Initial PE + KE

30 + 0.5mv^2 = 15 + 0.5mu^2

30 + 0.5*1.04*v^2 = 15 + 0.5*1.04*9^2

v = 7.22 m/s answer

b] force = - slope of the given curve

= - (15-30)/(4-2)

= 7.5 N in positive direction

c] Total mechanical energy is conserved,

Final PE + KE = Initial PE + KE

40 + 0.5mv^2 = 15 + 0.5mu^2

40 + 0.5*1.04*v^2 = 15 + 0.5*1.04*9^2

v = 5.74 m/s answer

b] force = - slope of the given curve

= - (40-15)/(6-5) = -25

= 25 N in negative direction answer