1. Fig 8-50 shows a plot of potential energy U versus position x of a 0.90 kg pa
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Question
1. Fig 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved). Three values are UA = 15.0j, UB = 35.0j, and UC =45.0j. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. (a) if the partcle can reach x = 1.0 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 left of x = 4.0m? Suppose instead, of the particle headed in thee positive x direction when it is released at x = 4.5m at speed 7.0m/s. (d) if the particle can reach x = 7.-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 = 5.0 m?Explanation / Answer
A) At x = 4.5 m potential energyis U1 = 15 J Initial kinetic energy isK1 = (1/2)mu2 m = 0.9 kg u= 7 m/s then K1 = (1/2)*0.9*(7)2J= 22.05 J Then the total energy is : T =15+22.05=37.05 J At x = 1m potential energy is U2 = 35J Let Kinitic energy be K2 Then total energy is : T = U2 +K2 = 37.05J 35 J+K2 = 37.05J Then K2 = 2.05 JThis is greater than zeroand hence particle reaches the point X=1m then K2 = (1/2)mv2 = 2.05J (1/2)*1.1(v)2J =2.05 J solve for v b)we know that force F = -dU/dX = -(35-15)/2-4 = +10 N c) As force ispositive then it points in the positiveX-direction. d) At X = 7m then potential energy is U3 = 45J Thisis greater than total energy 37.05J.Thus ball does not reach X = 7m .at turning point kinetic energy of theball is zero. then turning point will be between x = 5 mand x = 6 m total potential energy U4 is15-f(dx) here dx = x-5 we know that f = -dU/dX At X = 5 m dx = 6-5 = 1m dU = 45-15 = 30 J THEN f = -30/1 = -30N Then TOTAL POTENTIAL ENERGY IS U4 = 15+30(x-5) but total potential energy at turning point is equal tototalenergy 41.1855J then 15+30(x-5)= 41.1855J solve for x e) force at x = 5 m is f = -30 N F DIRECTION IS -VE X DIRECTION Hope this helps u! Then TOTAL POTENTIAL ENERGY IS U4 = 15+30(x-5) but total potential energy at turning point is equal tototalenergy 41.1855J then 15+30(x-5)= 41.1855J solve for x e) force at x = 5 m is f = -30 N F DIRECTION IS -VE X DIRECTION Hope this helps u!Related Questions
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