Question

A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30° (figure 6-E5). The friction coefficient between the road and the tyre is 1/2√3. 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.Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "

Answer 1

Solution :
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Hardest brake means maximum force of friction developed between car's tyre and road
Maximum frictional force=μR
From the free body diagram :
R-mg Cosθ=0
R=mg Cosθ------------1
and
μR+ma-mgsinθ=0-------------2
μmgcosθ+ma-mg sinθ=0
mgCosθ+a-10 x 1/2 =0
a=5-[1-2√3)]x10√3/2
=5-2.5=2.5 ms/2
s=12.8 m
u=6 m/s
Velocity at the end : 
V=√u²+2as
=√6²+2x2.5x12.8
=√36+64
=√100=10  m/s =36km/h

hence how hard the driver applies the brakes that car reaches the bottom with least velocity of 36km/hr

Answer 2

Answer:

Explanation:

Maximum frictional force=μR

From the free body diagram :

R-mg Cosθ=0

R=mg Cosθ------------1

and

μR+ma-mgsinθ=0-------------2

μmgcosθ+ma-mg sinθ=0

mgCosθ+a-10 x 1/2 =0

a=5-[1-2√3)]x10√3/2

=5-2.5=2.5 ms/2

s=12.8 m

u=6 m/s

Velocity at the end : 

V=√u²+2as

=√6²+2x2.5x12.8

=√36+64

=√100=10  m/s =36km/h