U A = 15 J, U B = 35 J and U C = 45 J. The particleis released at x = 4.5 m with
ID: 1726182 • Letter: U
Question
UA = 15 J,UB = 35 J andUC = 45 J. The particleis released at x = 4.5 m with an initialspeed of 6.90 m/s, headed in the negative x direction.(a) If the particle can reach x = 1.0 m,what is its speed there, and if it cannot, what is its turningpoint? What are the (b) magnitude and(c) direction (type 1 if positive direction and 2if negative one) of the force on the particle as it begins to moveto the left of x = 4.0 m? Suppose, instead, the particleis headed in the positive x direction when it is releasedat x = 4.5 m at speed 6.90 m/s. (d) Ifthe particle 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 (f) direction(type 1 if positive direction and 2 if negative one) of the forceon the particle as it begins to move to the right of x =5.0 m?
Explanation / Answer
The particle is released at x = 4.5 m The initial speed of the particle is u = 6.90 m/s (a)The kinetic energy of the particle at x = 4.5 m is equal tothe potential energy of the particle.Therefore,we get (1/2)mu2 = m * a * x or a = (u2/2x) ------------------(1) Let the speed of the particle when it reaches x = 1.0 m bev,therefore,we get v2 - u2 = 2a * (x - x1)-----------------(2) Here,x = x1= 1.0 m From equations (1) and (2),we get v2 - u2 = 2 * (u2/2x) * (x -x1) or v2 = u2 + 2 * (u2/2x) * (x- x1) or v2 = u2 + (u2/x) * (x -x1) or v2 = u2 * [1 + ((x -x1)/x)] or v2 = u2 * [1 + (1 -x1/x)] or v = u * [1 + (1 - x1/x)]1/2 or v = u * [2 - x1/x]1/2------------------(3) Substituting the values in the above equation,we get v = 6.90 * [2 - 1.0/4.5] = 12.267 m/s The speed of the particle at x = 1.0 m is v = 12.267m/s. (b and c)The magnitude and direction of the force on theparticle as it begins to move to the left of x = 4.0 m is F = m * a -----------------(4) Here,m = 1.14 kg and a = (u2/2x) =((6.90)2/(2 * 4.0)) = 5.95 m/s2 Substituting the values in equation (4),we get F = 1.0 * 5.95 = 5.95 N The direction of the force on the particle is towards theleft. (d)Let the speed of the particle when it reaches x = 7.0 m bev,therefore from equation (3),we get v = u * [2 - x1/x]1/2 Here,x = x1= 7.0 m or v = 6.9 * [2 - 7.0/4.5] or v = 3.067 m/s(e and f)The acceleration of the particle as it begins to moveto the right of x = 5.0 m is a = (v2/2x) or a = ((3.067)2/(2 * 5.0)) = 0.940m/s2
The magnitude and direction of the force on the particle as itbegins to move to the right of x = 5.0 m is F = m * a or F = 1.0 * 0.940 = 0.940 N The direction of the force on the particle is towards theright.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.