Question

Notify Me Bookmark

A conservative force F(x) acts on a 1.5 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is graphed in the figure. When the particle is at x = 2.0 m, its velocity is −1.4 m/s. (a) What is F(x) at this position, including sign? Between what positions on the (b) left and (c) right does the particle move? (d) What is its particle's speed at x = 7.0 m? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

A conservative force F(x) acts on a 1.5 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is graphed in the figure. When the particle is at x = 2.0 m, its velocity is −1.4 m/s. (a) What is F(x) at this position, including sign? Between what positions on the (b) left and (c) right does the particle move? (d) What is its particle's speed at x = 7.0 m? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

Image text
A conservative force F ( x ) acts on a 1.5 kg particle that moves along an x axis. The potential energy U ( x ) associated with F ( x ) is graphed in the figure. When the particle is at x = 2.0 m , its velocity is − 1.4 m / s . (a) What is F ( x ) at this position, including sign? Between what positions on the (b) left and (c) right does the particle move? (d) What is its particle's speed at x = 7.0 m ? (a) Number i Units ◻ (b) Number i ◻ Units ◻ (c) Number i Units ◻ (d) Number i Units ◻

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

The figure here shows a plot of potential energy U versus position x of a 0.884 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and UC = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.82 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 7.82 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? (a) Number Unit (b) Number Unit (c) (d) Number Unit (e) Number Unit (f)
Solution
0 0

The figure shows a plot of potential energy U versus position x of a 0.230 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.00 J, UC = 20.0 J and UD = 24.0 J. The particle is released at the point where U forms a "potential hill" of "height" UB = 12.0 J, with kinetic energy 6.00 J. What is the speed of the particle at (a)x = 3.50 m and (b)x = 6.50 m? What is the position of the turning point on (c) the right side and (d) the left side? (a) Number Units (b) Number Units (c) Number Units (d) Number Units
Solution
0 0

The figure here shows a plot of potential energy U versus position x of a 0.852 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and UC = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.89 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 7.89 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?
Solution
0 0