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13.A vehicle has 1000 kg mass and co-efficient of friction between tyres and road is 0.4what is the maximum breaking force (retarding force) which can stop the vehicle onroad-23924 Nb) 9180 Nc) 1000Nd) 400 N |
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Answer» Force due to friction is mgu where m is the mass, g is gravity, and I is the coefficient of friction. That gives you 1000*9.8*0.4 which equals 3920. hence , the correct option is (a) If you want the acceleration as well( I know you didn’t ask but I’m bored) you use F=ma. Solve for acceleration gives you a=F/m. 3920/1000=3.92. This means that the maximum acceleration that car can experience is 3.92m/s^2. This applies to speeding up and slowing down. |
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