The retarding acceleration of 7.35ms−2 due to frictional force stops the car of mass 400kg travelling on a road. The coefficient of friction between the tyre of the car and the road is
A
0.55
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B
0.75
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C
0.70
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D
0.65
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Solution
The correct option is B0.75 Using Newtons 2nd law Fnet=ma μmg=ma ⇒μ=mamg=ag
(a=7.35ms−2 given) ∴μ=7.359.8=0.75