The potential energy of a particle undergoing 1d motion...

The potential energy of a particle undergoing 1D motion along the y-axis is U(y) = 5cy2, where c = 2 N/m. Its total energy at y = 0 is 12 J, and it is only subject to conservative forces. Find a) the positions where its kinetic energy is zero. b) the forces at those positions.

A particle can move along only an x axis, where conservative forces act on it (see the figure and the table below). The particle is released at x = 5.00 m with a kinetic energy of K = 18.0 J and a potential energy of U = 0.0 J. If its motion is in the negative direction of the x axis, what are its (a) K and (b) U at x = 2.00 m and its (c) K and (d) U at x = 0? If its motion is in the positive direction of the x axis, what are its (e) K and (f) U at x = 11.0 m, its (g) K and (h) U at x = 12.0 m, and its (i) K and (j) U at x = 13.0 m? Next, the particle is released from rest at x = 0. What are (k) its kinetic energy at x = 5.0 m and (l) the maximum positive position xmax it reaches? Range | Force 0 to 2.00 m | F1 = +(3.00 N)i 2.00 m to 3.00 m | F2 = +(5.00 N)i 3.00 m to 8.00 m | F = 0 8.00 m to 11.0 m | F3 = -(4.00 N)i 11.0 m to 12.0 m | F4 = -(1.00 N)i 12.0 m to 15.0 m | F = 0

Priyanka K.