An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #135 KJ# to # 45 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2016-05-15T04:23:38

Average Speed #=136.6025404" "#m/sec

The given data:
Mass #m=9" "#kg
Kinetic Energy initial #KE_i=135000" "#joule
Kinetic Energy final #KE_f=45000" "#joule

initial time #t_0=0" "#sec
final time #t_1=4" "#sec

Compute for initial velocity #v_i#

#KE_i=1/2mv_i^2#

#135000=1/2(9)v_i^2#

#v_i^2=(2(135000))/9#

#v_i=sqrt(30000)#

#v_i=173.2050808" "#m/sec

Compute for final velocity #v_f#

#KE_f=1/2mv_f^2#

#45000=1/2(9)v_f^2#

#v_f^2=(2(45000))/9#

#v_f=sqrt(10000)#

#v_f=100" "#m/sec

Solve for the total distance traveled by the object using the following:

#v_f^2=v_i^2+2a*s" "#first equation
and
#v_f=v_i+a*t" "#second equation

#v_f^2-v_i^2=2a*s" "#from the first equation

#color(red)((v_f+v_i)(v_f-v_i)=2*a*s)#

#color(blue)((v_f-v_i)=a*t" ")#from the second equation

Divide the first equation by the second equation

#color(red)((v_f+v_i)cancel(v_f-v_i))/color(blue)(cancel(v_f-v_i))=color(red)(2*cancela*s)/color(blue)(cancela*t)#

#v_f+v_i=(2s)/t#

#s=1/2*t(v_f+v_i)#

#s=1/2*(4)(100+173.2050808)#

#s=546.4101616" "#m

Average speed #=("distance traveled")/("elapsed time")#

Average speed #=546.4101616/4#

Average speed #=136.6025404" "#m/sec

God bless....I hope the explanation is useful.