Question

4. Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a

runaway 60,000 kg truck is moving at 27 m/s, how much work must be done to stop the truck?

Answer 1

21870 kJ

Step-by-step explanation:

Kinetic energy of the truck is equivalent to the work required to stop it

`KE=0.5mv^(2)` where m is the mass of truck while v is the speed which it moves with. Substituting 60000 kg for m and 27 m/s for v then

`KE=0.5*60000*27^(2)=21870000 J= 21870 kJ`