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प्रश्न
A care is going at a speed of 21.6 km/hr when it encounters at 12.8 m long slope of angle 30° (in the following figure). The friction coefficient between the road and the tyre is `1/2sqrt3`. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s2.
उत्तर
When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.
So, maximum frictional force = μR
From the free body diagram,
R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
μR + ma − mg sin θ = 0 (2)
⇒ μ mg cos θ + ma − mg sin θ = 0
where θ = 30˚
`=>mug cos theta+a-10xx(1/2)=0`
`=>a=5-[1/2sqrt3]xx10(sqrt3/2)`
`=5-(10xx3)/4`
`=(-10)/4`
= -2.5 m/s2
s = 12.8 m
u = 6 m/s
∴ Velocity at the end of incline
`v=sqrt(u^2+2as)`
`=sqrt(6^2+2(2.5)(12.8))`
`=sqrt(36+64)`
= 10 m/s = 36 km/h
Therefore, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h.
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