\"a .20kg particle moves along the x axis under the influenceof a stationary obj

Question

"a .20kg particle moves along the x axis under the influenceof a stationary object. the potential energy is given by U(x)= 8x^2+2x^4 , where U is in Joules and X is in meters. if theparticle has a speed of 5.0 m/s when it is at x= 1.0 m , it'sspeed when it is at the orgin is : A) 0 B) 2.5 m/s C) 5.7 m/s D) 7.9 m/s E) 11 m/s "a .20kg particle moves along the x axis under the influenceof a stationary object. the potential energy is given by U(x)= 8x^2+2x^4 , where U is in Joules and X is in meters. if theparticle has a speed of 5.0 m/s when it is at x= 1.0 m , it'sspeed when it is at the orgin is : A) 0 B) 2.5 m/s C) 5.7 m/s D) 7.9 m/s E) 11 m/s

Explanation / Answer

Energy must be conserved. So, sum of Kinetic and potential energy at x=1 m must be equal tothat at x=0. KE at x=1 = (1/2)mv2 = (1/2)(.2)(5)2 = 2.5J P.E = U = 8x^2+2x^4 = 8(1)2 + 2(1)4 = 10J So, total energy = 12.5 J At x=0, PE = U = 8x^2+2x^4 = 8(0)2 + 2(0)4 =0 J Hence, KE must be 12.5 J because PE+KE must be same as at x=1m So, (1/2)mv2 = (1/2)(0.2)v2= 12.5 or v= 11.18 m/s Hence, answer is E.

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