contestada

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?

Respuesta :

opudodennis

Answer:

21870 kJ

Explanation:

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

[tex]KE=0.5mv^{2}[/tex] 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

[tex]KE=0.5*60000*27^{2}=21870000 J= 21870 kJ[/tex]

Otras preguntas