E DA 0.20-kg particle moves along the x axis under the influence of a conservative force. The potentialenergy is given byUx) = (8.0 J/m²)x² + (2.0 J/m*)x*,where x is in coordinate of the particle. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, itsspeed when it is at the origin is:A) 0 m/sB) 2.5 m/sC) 5.7 m/sD) 7.9 m/s(Е) 11 m/s

Answered: Image /qna-images/answer/644e9357-08e4-4db5-a2e1-6b771c6d9e00.jpg

1DA 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by Ux) = (8.0 J/m³)x² + (2.0 J/m*)x", where x is in coordinate of the particle. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is: A) 0 m/s B) 2.5 m/s C) 5.7 m/s 7.9 m/s 11 m/s

1DA 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by Ux) = (8.0 J/m³)x² + (2.0 J/m*)x", where x is in coordinate of the particle. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is: A) 0 m/s B) 2.5 m/s C) 5.7 m/s 7.9 m/s 11 m/s

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

only along an x axis under the influence of a conservative force. On the graph UA 9.00 J, Uc 20.0 J, and Up = 24.0 J. The particle is released at the point where UB = 12.00 J with kinetic energy 4.00 J. Which of the following statements must be true? = tot Uch ŜUR UA 0 6 7 a. If the particle is initially moving in the −î direction, then it has enough energy to reach x = 0. The particle is at an unstable equilibrium at x = 5.00 m. b. c. The kinetic energy of the particle is greater than 4.00 J at x = 6.50 m. 7.50 m. d. The work done on the particle must be negative as it moves from x = 7.00 m to x = e. The particle will be stuck in the potential well between x = 6.00 m and x = 7.00 m.
A box is given a push across a horizontal surface. The box has a mass m, the push gives it an initial speed of 2.40 m/s, and the coefficient of kinetic friction between the box and the surface is 0.140. (a) Use energy considerations to find the distance (in m) the box moves before it stops. m (b) What If? Determine the stopping distance (in m) for the box if its initial speed is doubled to 4.80 m/s. m
A variable force F = kx² acts on a particle which is initially at rest. Calculate the work done by the force during the displacement of the particle from x = 0 m to x = 7 m. (Assume the constant k =3 N m-2)
In heavy rush-hour traffic, you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5 minutes, and finally, you drive at 15 m/s for another 2.5 minutes. Plot a position-versus-time graph for this motion. Your plot should extend from t = 0 minutes to t = 7.5 minutes. Assume x=0 and t=0 at the start of your motion.
None
The figure here shows a plot of potential energy U versus position x of a 0.888 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are Ug = 15.0 J, Ug = 35.0 J and Uc = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.04 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.04 m/s. (d) If the particle can reach x = 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.00 m? Uc UB UA 4 6 x (m) (a) Number Unit (b) Number Unit (c) (d) Number…
show work
A flywheel in the form of a uniformly thick disk of radius 1.53 m has a mass of 58.6 kg and spins counterclockwise at 141 rpm. Calculate the constant torque required to stop it in 3.25 min. N.m constant torque:
The figure below shows a plot of potential energy U versus position x of a 0.76 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are UA = 20.0 J, UB = 35.0 J, and Uc = 45.0 J. Uc UB UA 4 х (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? m (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? magnitude direction +X 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? m (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? magnitude…
A 9.00-kg particle moves from the origin to position ©, having coordinates x = 4.50 m and y = 4.50 m as shown in the figure. One force on the particle is the gravitational force acting in the negative y direction. Using the equation (W = FAr cos e = F . Ař), calculate the work done by the gravitational force in going from O to © along the following paths. у (m) C) (x, y) х (m) (a) the purple path (o0© (b) the red path (OB© (c) the blue path (O© (d) Your results should all be identical. Why?
The figure here shows a plot of potential energy U versus position x of a 0.878 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA-15.0J, U₂ - 35.0J and U-45.0 J. The particle is released atx -4.50 m with an initial speed of 7.99 m/s, headed in the negative x direction. (a) If the particle can reach x - 1.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x-4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x -4.50 m at speed 7.99 m/s. (d) If the particle can reach x - 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x-5.00 m? U(J) Uc UB UA 2 4 x (m) 6
A baby is free to move on a horizontal frictionless table and is attracted to the origin by a mysterious force with magnitude F(r) = ar+Br². (i) Find the potential energy function associated with the mysterious force. (ii) If the baby is released from rest at the location (r, 0) = (a, π/4), what will be his speed when he reaches the origin? Law Application a X