Answer:
By work-energy theorem, Loss in K.E. of the vehicle \[=\]Work done against the force of friction x distance or K.E. \[=f\times s=\mu R\times s=\mu \text{mg s}\] or \[s=\frac{K.E.}{\text{ }\!\!\mu\!\!\text{ mg}}\] For constant K.E., \[s\propto \frac{1}{m}\] As the truck has more mass than the car, so it will stop in a lesser distance than the car.
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