A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time, (d) the net torque about the origin exerted on the particle as a function of time,(e)the angular momentum of the particle as a function of time,(f)the kinetic energy of the particle as a function of time, and (h)the power injected into the system of the particle as a function of time.Hint: (a) v = dr/dt and dr = vdt, r =òvdt (integrate v and get r),(b) a=dv/dt, © F=ma,(d) t = r x F (cross product) (e) L= r x mv(cross product) (f) KE=(1/2)mv2(Square v vector to get v2,i2 =j2 =1),(g) Power = first derivative of KE

Solution given thatm=5 kgu=6t2i +2tjms

Answered: A 5.00-kg particle starts from the…

An Introduction to Physical Science

14th Edition

ISBN:9781305079137

Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Chapter2: Motion

Section2.3: Acceleration

Problem 2.3CE: What is the average speed in mi/h of a person at the equator as a result of the Earths rotation?...

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A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time, (d) the net torque about the origin exerted on the particle as a function of time,(e)the angular momentum of the particle as a function of time,(f)the kinetic energy of the particle as a function of time, and (h)the power injected into the system of the particle as a function of time.

Hint: (a) v = dr/dt and dr = vdt, r =òvdt (integrate v and get r),(b) a=dv/dt, © F=ma,(d) t = r x F (cross product) (e) L= r x mv(cross product) (f) KE=(1/2)mv2(Square v vector to get v2,i2 =j2 =1),(g) Power = first derivative of KE

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