Question

The figure below shows a plot of potential energy U versus position x of a 1.36 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 15), U2 = 30 ), and U3 = 40 ). Ur--- - U (1) --- + - --- 2 4 6 x (m) The particle is released at x = 4.5 m with an initial speed of 6.0 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 particle as it begins to move to the left of x = 4.0 m? N in the ---Select--- direction. Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 6.0 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? ---Select--- (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? N in the ---Select--- y direction.

Answer 1

TE = KE & PE at 2 = 4.5m, TE = (3x196x6²) + u TE = 24.48+ is = 39.48 J at x=LM, I PE = U2 = 303 <TE KE = TE-PE = 39.48 - 30 1x 1.364 v² = 9.48 Tv = 3.72 m18 6 f= - 20 = - (slope) = - ( 15 - 20 N = 7.50 to the right (the s-axis) 6 s TE, so it cannot reach there. Umax = 39.48 = (40-10) (2-5) +30 [x = 5.96m turning point a nas, F= -( 40-15) = – 250 Anes 25N, along -ve x-axis Are! -ve x-axis