The work done on the truck to achieve given speed is [tex]7.62*10^{6}[/tex] Joules.
The work done on the truck is equivalent to the kinetic energy.
Work done = Kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex]
Where m is mass of truck and v is final velocity of truck.
Given that, [tex]m=3120Kg,v=22.1m/s[/tex]
Substitute values in above equation
[tex]Workdone=\frac{1}{2}*3120*(22.1)^{2} \\\\Workdone=0.5*3120*488.41=7.62*10^{6} KJ[/tex]
Thus, The work done on the truck to achieve given speed is [tex]7.62*10^{6}[/tex] Joules.
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