Question

Chapter 08, Problem 077 Your answer is partially correct. Try again. A conservative force F(x) acts on a 1.9 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 -1.3 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? x (m 10 5 15 -5 3-10 -15 20 (a) Number Units 4.9 N (b) Number T1.75 Units m (c) NumberT13.25 Units m Units Tm/s (d) Number 3.341209992 (r) (*

Answer 1

© © - Mass of partice, malog kg At 22=2.0m , velou try V=-1.3 m/s. Since, Ferdafe --(-7-22 ri FCO)? -|-4.7] N FCU- 4.7 N At = 2:00 U=-7.01 Using conservahon of energy the Utca Vitki 8-70674, X1:921-13335 Ufs- 5.9941 Now, funchong Ule) is U(X)= -3 -136-1) 407 forror <4 . for j = Ube) -5.9943 -3 -6347) 4.7 >> (+)4.7 = -3 +5.994 > >=14647 @