Question

The figure here shows a plot of potential energy U versus position x of a 0.886 ko partide that can travel only along an x ads.Noncordervative forces are not involved.) Three values are U 1 5), Us 35 ) and UC - 45 ). The partide is released at x - 4.5 m with an initial speed of 7.8 m/s, headed in the negative x direction. (a) If the partide can reach x™ 1.0 m, what is its speed there, and it it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the partides begins to move to the left of x 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x 4.5 m at speed 7.6 ms. (d) the partide can reach x 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (1) direction of the force on the partide as it begins to move to the right of x = 5.0 m? Unit (a) Number (b) Number (c) Direction (d) Number (e) Number 10:24 PM

Answer 1

At x= 4.5m, i15J the speed v .8ms them the Kinetc enengy is n o Kg)ms) 26-95 J 495J Total emergy EU+Ki a At x1 m, V2 35J By energy Consewottm 6.95J0 kr= u95J-35 we have 2 6-953) 04896kg 2.96 mls (353-153-10N (2-4)m 4U the brace points in the +x divecim 'means, d) At xm, V3 -u5T, cohich exceeds the initia total enengy (Fi).Thus, the particle Can nevev eaeh there At the tuonina port ks =o.Retoeen xSm & 6m 41e potenal emergy is given by vA)=15 +30-s), 5x Thus, u.953=15+30(x-)xs.898m (us-15)3 (6-5) m eAt x 5m, Fr =-40 -30N () 30N 4) キx<0, the brace poins n he-xdirectem