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Numericals · Q30

Q.A truck of mass 5 ton is travelling on a horizontal road with 36 km/hr and stops on travelling 1 km after its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton, i.e., a car moving with same speed stops in similar distance, how much will the fraction be? (Use g = 10 m/s^2, unless otherwise stated.)

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Truck: mass 5 ton, u=36u=36 km/hr =10=10 m/s, stops (v=0v=0) after s=1s=1 km =1000=1000 m. From v2=u2−2asv^2=u^2-2as: 0=100−2a(1000)⇒a=1002000=0.050=100-2a(1000)\Rightarrow a=\frac{100}{2000}=0.05 m/s^2. The friction force is F=maF=ma, and its fraction of the weight mgmg is Fmg=mamg=ag=0.0510=1200\frac{F}{mg}=\frac{ma}{mg}=\frac{a}{g}=\frac{0.05}{10}=\frac{1}{200}. For the car (mass 1 ton, same initial speed 10 m/s, same stopping distance 1000 m, same road/surface): the calculation for a is IDENTICAL (a depends only on u and s, not on mass), so a=0.05a=0.05 m/s^2 again, and the fraction ag=1200\frac{a}{g}=\frac{1}{200} again -- the mass cancels out of the fraction entirely, so the SAME fraction (1/200) applies regardless of the vehicle's mass, as long as u and s (and hence a, and the coefficient of friction) are the same. [!ANSWER] The frictional force is 1/200th of the vehicle's weight, for BOTH the 5 ton truck and the 1 ton car -- the fraction is mass-independent since it depends only on a/ga/g, and a itself depends only on u and s (same for both vehicles here), not on mass.

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