Ejercicio Exercise 1 · Física I URV 2024

Tema: Treball, energia i energia potencial · Del examen Physics I - 1st & 2nd Partial Exam (January 2024) (2024)

Enunciado

Exercise 1 (10 points). Potential energy, conservative force, and 1D motion

Consider a particle moving in one dimension under the action of a conservative force characterized by the potential energy function:

U(x)=9x2−16U(x) = 9x^2 - 16

where position xx is given in centimeters (cm) and the potential energy U(x)U(x) in Joules (J).

a)

Does the coefficient of x2x^2 have physical units? If so, state them. [1 point]

b)

Determine the interval of movement allowed for the particle if its total mechanical energy is Em=6 JE_m = 6\text{ J}. Sketch the kinetic energy curve Ek(x)E_k(x) strictly within this physically allowed interval. [2 points]

[

,

] cm

c)

Plot the force F(x)F(x) acting on the particle as a function of position within this same interval of motion. [2 points]

F(x)=F(x) =

J/cm

d)

Repeat questions b) and c) assuming that the total mechanical energy is instead Em=−5 JE_m = -5\text{ J}. [4 points]

[

,

] cm

e)

What is the minimum mechanical energy required for the particle to start moving? Justify your answer. [1 point]

Em>E_m >

J

Enunciado: Ejercicio Exercise 1 · Física I URV 2024

Solución

Según los criterios oficiales de corrección

a. Does the coefficient of x² have physical units? If so, state them.

Solución: J/cm2\text{J/cm}^2 / J/m2\text{J/m}^2 / N/cm\text{N/cm} / N/m\text{N/m}

b. Determine the lower turning point x_min (cm) for Em = 6 J

Solución: −223-\frac{\sqrt{22}}{3} cm / −1,563-1{,}563 cm

b. Determine the upper turning point x_max (cm) for Em = 6 J

Solución: 223\frac{\sqrt{22}}{3} cm / 1,5631{,}563 cm

c. Expression of the force F(x) acting on the particle

Solución: −18x-18x

d. Lower turning point x_min (cm) for Em = -5 J

Solución: −113-\frac{\sqrt{11}}{3} cm / −1,106-1{,}106 cm

d. Upper turning point x_max (cm) for Em = -5 J

Solución: 113\frac{\sqrt{11}}{3} cm / 1,1061{,}106 cm

e. Minimum mechanical energy required for the particle to start moving

Solución: −16-16

e. Justification of the minimum mechanical energy required

Solución: At x=0x = 0, U(x)U(x) reaches its minimum U(0)=−16 JU(0) = -16\text{ J}. At Em=−16 JE_m = -16\text{ J}, Ek=0E_k = 0 and the particle is in static equilibrium; for motion to occur, Ek>0E_k > 0, so Em>−16 JE_m > -16\text{ J}.

Sketches and Plots: b) Ek(x) curve for Em = 6 J; c) F(x) plot for Em = 6 J; d) Ek(x) and F(x) plots for Em = -5 J

Solución: b. Sketch of Ek(x)E_k(x) for Em=6 JE_m = 6\text{ J}: Inverted parabola symmetric around x=0x = 0 with a maximum of Ek(0)=22 JE_k(0) = 22\text{ J}, dropping to 0 J0\text{ J} at the turning points x=±223 cm≈±1,563 cmx = \pm\frac{\sqrt{22}}{3}\text{ cm} \approx \pm 1{,}563\text{ cm}. Strictly plotted inside this interval. c. Plot of F(x)F(x): Straight line passing through the origin with negative slope −18 J/cm2-18\text{ J/cm}^2 (or −1800 N/cm-1800\text{ N/cm}). At x=−1,563 cmx = -1{,}563\text{ cm}, F=+28,14 J/cm=+2814 NF = +28{,}14\text{ J/cm} = +2814\text{ N}; at x=0x = 0, F=0F = 0; at x=+1,563 cmx = +1{,}563\text{ cm}, F=−28,14 J/cm=−2814 NF = -28{,}14\text{ J/cm} = -2814\text{ N}. d. Plots for Em=−5 JE_m = -5\text{ J}: Ek(x)=11−9x2E_k(x) = 11 - 9x^2 is an inverted parabola reaching peak Ek(0)=11 JE_k(0) = 11\text{ J} and vanishing at x=±113 cm≈±1,106 cmx = \pm\frac{\sqrt{11}}{3}\text{ cm} \approx \pm 1{,}106\text{ cm}. The force F(x)=−18xF(x) = -18x is identical in functional form but plotted over the narrower domain [−1,106 cm,1,106 cm][-1{,}106\text{ cm}, 1{,}106\text{ cm}], spanning from +19,9 J/cm+19{,}9\text{ J/cm} to −19,9 J/cm-19{,}9\text{ J/cm}.

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