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Check Your Understanding Find x ( t ) for the mass-spring system in [link] if the particle starts from x 0 = 0 at t = 0 . What is the particle’s initial velocity?

x ( t ) = ± ( 2 E / k ) sin [ ( k / m ) t ] and v 0 = ± ( 2 E / m )

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Summary

  • Interpreting a one-dimensional potential energy diagram allows you to obtain qualitative, and some quantitative, information about the motion of a particle.
  • At a turning point, the potential energy equals the mechanical energy and the kinetic energy is zero, indicating that the direction of the velocity reverses there.
  • The negative of the slope of the potential energy curve, for a particle, equals the one-dimensional component of the conservative force on the particle. At an equilibrium point, the slope is zero and is a stable (unstable) equilibrium for a potential energy minimum (maximum).

Problems

A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.

10 x with x -axis pointed away from the wall and origin at the wall

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A single force F ( x ) = −4.0 x (in newtons) acts on a 1.0-kg body. When x = 3.5 m, the speed of the body is 4.0 m/s. What is its speed at x = 2.0 m?

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A particle of mass 4.0 kg is constrained to move along the x -axis under a single force F ( x ) = c x 3 , where c = 8.0 N/m 3 . The particle’s speed at A , where x A = 1.0 m, is 6.0 m/s. What is its speed at B , where x B = −2.0 m?

4.6 m/s

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The force on a particle of mass 2.0 kg varies with position according to F ( x ) = −3.0 x 2 ( x in meters, F ( x ) in newtons). The particle’s velocity at x = 2.0 m is 5.0 m/s. Calculate the mechanical energy of the particle using (a) the origin as the reference point and (b) x = 4.0 m as the reference point. (c) Find the particle’s velocity at x = 1.0 m . Do this part of the problem for each reference point.

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A 4.0-kg particle moving along the x -axis is acted upon by the force whose functional form appears below. The velocity of the particle at x = 0 is v = 6.0 m/s . Find the particle’s speed at x = ( a ) 2.0 m , ( b ) 4.0 m , ( c ) 10.0 m , ( d ) Does the particle turn around at some point and head back toward the origin? (e) Repeat part (d) if v = 2.0 m/s at x = 0 .

A graph of F of x, measured in Newtons, as a function of x, measured in meters. The horizontal scale runs from 0 to 8.0, and the vertical scale from-10.0 top 10.0. The function is constant at -5.0 N for x less than 3.0 meters. It increases linearly to 5.0 N at 5.0 meters, then remains constant  at 5.0 for x larger than 5.0 m.

a. 5.6 m/s; b. 5.2 m/s; c. 6.4 m/s; d. no; e. yes

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A particle of mass 0.50 kg moves along the x -axis with a potential energy whose dependence on x is shown below. (a) What is the force on the particle at x = 2.0 , 5.0 , 8.0 , and 12 m? (b) If the total mechanical energy E of the particle is −6.0 J, what are the minimum and maximum positions of the particle? (c) What are these positions if E = 2.0 J? (d) If E = 16 J , what are the speeds of the particle at the positions listed in part (a)?

The energy I of x in Joules is plotted as a function of x in meters. The horizontal scale runs from less thqan zero to over 20 meters, but is labeled only from 0 to 20. The vertical scale runs from –12.0 to 12 Joules. U of x is a cponstant 4.0 Joules for all x less than 4.0 meters. It rises linearly to 12.0 Joules at 6.0 meters, then decreases linearly to –12.0 Joules at 10.0 meters. It remains –12.0 Joules from 10.0 to 14.0 meters, then rises to 12.0 Joules at 18 meters. It remains at 12.0 joules for all x larger than 18 meters.
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(a) Sketch a graph of the potential energy function U ( x ) = k x 2 / 2 + A e α x 2 , where k , A , and α are constants. (b) What is the force corresponding to this potential energy? (c) Suppose a particle of mass m moving with this potential energy has a velocity v a when its position is x = a . Show that the particle does not pass through the origin unless

A m v a 2 + k a 2 2 ( 1 e α a 2 ) .
The potential energy function U of x equal to k x squared over two plus A e to the alpha x squared is plotted as a function of x, with k=0.02, A=1, and alpha equal to one. The horizontal scale runs from –25 to 25 and the vertical scale runs from 0 to 4.5. The function is an upward opening parabola with a small Gaussian upward bump at the center. For the parameters chosen in this plot, the bump has a maximum value of one.

a. where k = 0.02 , A = 1 , α = 1 ; b. F = k x α x A e α x 2 ; c. The potential energy at x = 0 must be less than the kinetic plus potential energy at x = a or A 1 2 m v 2 + 1 2 k a 2 + A e α a 2 . Solving this for A matches results in the problem.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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