IIT JEE , AIEEE Forum - Physics , Mechanics

elessar_iitkgp

The brakes only serve to lock the wheels. Now, the main battle is between gravity and friction. The car will move down if mgsin

mgcos

That is, if tan

Which is true in the given case. (Substitute the values and check)

Now, from the FBD,
a = g(sin

-

cos

)
v(dv/dx) = g(sin

-

cos

)
Integrating the above equation,
v² = v₀² + 2g(sin

-

cos

)L
Substitute v₀ = 21.6 km/h = 6 m/s, g = 10m/s² ,

= 1/(2

3) , L = 12.8 m
v = 10 m/s = 36 km/h

So irrespective of how hard the driver applies brakes (which merely locks the wheels), the car attains a speed of 36 km/h.

PS: Note that the above equation can be obtained by energy conservation, but as you have asked from friction chapter, its possible that you haven't read upto energy conservation. So I haven't used it.

Also note that if rolling motion of wheels is present, the friction between the tyres and the incline will decrease. In that case, the velocity at the bottom of the inclined plane increases.

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