Chapter 08, Problem 077- 3rd TIME ASKING SAME QUESTION!!!! A conservative force

Question

Chapter 08, Problem 077- 3rd TIME ASKING SAME QUESTION!!!!

A conservative force F(x) acts on a 2.0 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 –1.6 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?

                                                                                                                      ******B) is NOT 1.5 m

                                                                                                                     ******C) is 13.7 m

                                                                                                                      ******D) is 3.4 m/s

Explanation / Answer

slope of curve at (x=2 m) = dU/dx = (-17 - (-3))/(4-1) = -14/3

A) F(x) = -dU/dx = 14/3 = 4.67 N

accelartion at (x=2) = 4.67/2 = 2.33 m/s2

B) in left particle speed is zero at, v + at = 0 => -1.6 + 2.33t = 0 => t = 1.6/2.33 = 0.6867 s

s = ut + 0.5at2 = -1.6 x 0.6867 + 0.5 x 2.33 x 0.68672 = -0.5494 m

left position = 2 - 0.5494 = 1.4506 m

C) speed at (X=4), = sqrt((-1.6)2 + 2 x 2.33 x 2) = 3.45 m/s

speed at (x = 10) = 3.45 m/s

after 10, dU/dx = (-2 - (-17))/(15-10) = 3

accelaration = -3/2 = -1.5 m/s2

s' = 3.452/(2 x 1.5) = 3.96 m

so right distance = 13.96 m

D) speed at (X= 7) = 3.45 m/s

Note: little fluctuations may be due to graph value are not that accurately clear. Method is exact.

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