An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #72 KJ# to # 120 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

1 Answer
Mar 16, 2016

Answer:

#v_a=8/3 m/s#

Explanation:

#Delta E_k=E_"k l"-E_"k i"" " "changing of the kinetic energy of object "#
#1/2m(v_l^2-v_i^2)=120-72#
#1/2*cancel(6)(v_l^2-v_i^2)=cancel(48)#
#(v_l^2-v_i^2)=16#
#(v_l-v_i)(v_l+v_i)=16#
#v_l-v_i=2#
#v_l+v_i=8#
#2*v_l=10" "v_l=5 m/s#
#5-v_i=2#
#v_i=3 m/2#
#a=(Delta v)/(Delta t)=(v_l-v_i)/(Delta t)=(5-3)/(3-0)=1 m/s^2" (acceleration of object)"#
#v_l^2=v_i^2+2*a*Delta x#
#5^2=3^2+2*1*Delta x#
#25=9+2*Delta x#
#16=2*Delta x#
#Delta x=8 m" displacement for (0-3)"#
#"Average velocity of an object is given by:"#
#v_a=("Total displacement")/("Total time")#
#v_a=8/3 m/s#