D LM07-CB-Rota × rse Materi x \\ D Microsoft Wor. \"A HW09-WorkEn Secure https:/

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D LM07-CB-Rota × rse Materi x D Microsoft Wor. "A HW09-WorkEn Secure https://www.webassign.net/web/Student/Assignment-Responses/su.. Howdy zy Texas A&M; UniversitMy eCampus-Black My Notes Ask Your Tea s Answers OSUniPhys1 8.4.WA.062 Tutorial A particle with a mass m = 2.00 kg is moving along the x axis under the influence of the potential energy function U(x) (2.00 3/m2)x2 -32.0J. If the particle is released from rest at the position x (The sign is important. Be sure not to round Intermediate calculations.) 6.60 m, determine the following. (a) total mechanical energy of the particle at any position 55.12 (b) potential energy of the particle at the position x 4.80 m 14.08 (c) kinetic energy of the particle at the position x 4.80 m 41.08 (d) maximum speed of the particle as it travels along the x axis 7.424 The particle will have its maximum speed when the kinetic energy has its maximum value. How can we determine the maximum value for the kinetic energy? m/s (e) acceleration of the particle at the position x - 4.80 m (Indicate the direction with the sign of your answer.) m/s2 (f) maximum position the particle may obtain along the x axis

Explanation / Answer

d) using conservation of energy

-32+0.5 mv^2= 55.12

v= 9.33 m/s

e) force is given by

F= -dU/dx= -4x

at x= 4.8 m

F= -4.8*4= - 19.2

2*a= -19.2

a= -9.6 m/s/s

f) for max displacement, the potential energy must be maximum, so

2x*x-32= 55.12

x= 6.6 m

comment in case any doubt

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