Energy diagrams. 1. A particle of mass m = 2 kg moves in a region of space with the following potential energy function.

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1 Energy diagrams 1. A particle of mass m = 2 kg moves in a region of space with the following potential energy function U (J) x (m) a. The force at x = 2 m is (positive/negative/zero). In the interval x = [4,5] m, the force is (positive/negative/zero). Explain. b. Estimate the magnitude of the force at x = 5 m. c. Find the work done by the force when the particle moves from x = 4 m to x = 5 m. d. Is the sign of your answer to part c consistent with your answer to part a? e. Where is the force equal to zero?

2 The particle is initially at x = 11 m and has a velocity v = 2 5 m/s i. f. Find the total mechanical energy of the particle. g. Where is the speed of the particle minimum and maximum? h. Describe qualitatively the motion of the particle. i. Now imagine that the total mechanical energy of the particle is E = 40 J. Describe the motion of the particle. Where is the speed of the particle maximum/minimum? Determine these speeds. j. x = 6 m is an unstable equilibrium point. Explain why. Answers: 1. a) Positive. Negative. b) 30 N c) 30 J e) 3 m, 6 m, 10 m f) 20 J g) x = 10 m; vmin x = 7 m and x = 12 m (turning points) h) Oscillations between 7 and 12 m i) Oscillations between 1 and 12.5 m. vmax (=7.7 x = 10 m; vmin x = 1 m and x = 12.5 m (turning points) j) F = 0, but for any displacement there will be a force that pulls the particle away from this position.

3 Energy diagrams 1. A particle of mass m = 2 kg moves in a region of space with the following potential energy function (part b) (part i) (parts f, g, h) U (J) x (m) a. The force at x = 2 m is _positive (positive/negative/zero). In the interval x = [4,5] m, the force is negative (positive/negative/zero). Explain. The force is minus the slope of the U(x) curve at each point. b. Estimate the magnitude of the force at x = 5 m. The slope of the U(x) curve at x = 5 m is the slope of the tangent line (in blue the figure). That line goes through points (4 m, -10 J) and (5 m, 20 J), so the force is: U 20 ( 10) J F 30 N x 5 4 m c. Find the work done by the force when the particle moves from x = 4 m to x = 5 m. W U U (5 m) U (4 m) 20 ( 10) J 30 J d. Is the sign of your answer to part c consistent with your answer to part a? Yes: if you move from 4 m to 5 m (displacement in the +x direction) and the force is in the x direction, the work should be negative. e. Where is the force equal to zero? Whenever the slope of the curve is zero: at x = 3 m, 6 m and 10 m. The particle is initially at x = 11 m and has a velocity v = 2 5 m/s i.

4 f. Find the total mechanical energy of the particle. Initial potential energy (at x = 11 m): U = 0 J 2 2 Initial kinetic energy: K 1 mv 1 (2 kg)(2 5 m/s) 20 J 2 2 Mechanical energy: E = K + U = 20 J The mechanical energy in these conditions is the red horizontal line in the figure. g. Where is the speed of the particle minimum and maximum? The speed and the kinetic energy are minimum (maximum) whenever the potential energy is maximum (minimum) within the region allowed by the mechanical energy. This last part means that in our case the particle will move between x = 7 m and x = 12 m only. So the maximum speed happens at x = 10 m: K E U (10 m) 20 ( 20) 40 J 1 2K 2(40 J) 6.3 m/s 2 m (2 kg) 2 K mv v And the minimum speed happens at the two turn-around points where the speed is zero, x = 7 m and x = 12 m. h. Describe qualitatively the motion of the particle. The particle oscillates between x =7 m and x = 12 m. The particle has zero speed at the turn around points and is subject to a force that pulls it toward the minimum at x = 10 m. So the particle speeds up as it moves from the turn around point to the minimum and then slows down as it travels from the minimum to the other turn around point. i. Now imagine that the total mechanical energy of the particle is E = 40 J. Describe the motion of the particle. Where is the speed of the particle maximum/minimum? Determine these speeds. Now the system oscillates between x = 1 m and x = 12.5 m, but the oscillations are more complex than before, because there are two minima. If the particle begins at x = 1 m, for instance, it will speed up until x = 3 m, slow down from there to x = 6 m, speed up again from x = 6 m to x = 10 m, and then slow down and come momentarily to a stop at x = 12.5 m. It then begins moving in the negative x direction in a similar way. The speed of the particle is maximum at the lowest minimum of the potential energy function, ie at x = 10 m. There,

5 K E U (10 m) 40 ( 20) 60 J 1 2K 2(60 J) 7.7 m/s 2 m (2 kg) 2 K mv v The minimum speed happens at the two turn-around points where the speed is zero, x = 1 m and x = 12.5 m. j. x = 6 m is an unstable equilibrium point. Explain why. The slope of the curve at x = 6 m is zero, ie F = 0. So if we were to place a particle at rest at x = 6 m, it would stay there, so it is an equilibrium position. However, the smallest displacement to the left or to the right of x = 6 m would put the particle under the effect of a force to the left or right (respectively) that would pull the particle away from the equilibrium point. This happens because x = 6 m is a local maximum of the U(x) curve. Therefore, x = 6 m is an unstable equilibrium point.