Answer:[tex]5.43\times 10^5 J[/tex]
Explanation:
Given
mass of truck(m)=1330 kg
height of hill(h)=16.1 m
work done by friction[tex](W_r)=-3.23\times 10^5 J[/tex]
Work done by engine[tex](W_e)=6.57\times 10^5 J[/tex]
and
change in Kinetic energy=
[tex]\Delta K E+\Delta U_{g}=W_r+W_e[/tex]
[tex]\Delta K E=-3.23\times 10^5+6.57\times 10^5-1330\times 9.8\times 16.1[/tex]
[tex]\Delta K E=(-3.23+6.57+2.09)\times 10^5[/tex]
[tex]\Delta K E=5.43\times 10^5 J[/tex]