Question

a vehicle of mass M is moving on a rough horizontal Road with kinetic energy E. if the coefficient of friction between the tires Road being (mu), then the stopping distance is.
with clear cut explaination
please please​

Answer 1

Answer:

Initial velocity

u

=

p

m

Final velocity

v

=

0

, (as the vehicle must stop)

Force of friction

=

μ

m

g

(where

g

is acceleration due to gravity.)

Acceleration due to friction

=

−

μ

m

g

m

=

−

μ

g

(

−

ve sign shows that it is retardation.)

Using the kinematic expression

v

2

−

u

2

=

2

a

s

and inserting various values we get stopping distance

s

(

0

)

2

−

p

2

m

2

=

2

(

−

μ

g

)

s

⇒

s

=

p

2

2

m

2

μ

g

Answer 2

Answer:

$\frac{2E}{\mu mg}$

Explanation:

We are given that

Mass of vehicle=M

Initial K.E=E

Coefficient of friction=$\mu$

We have to find the value of stopping distance.

$KE_i=\dfrac{1}{2}mu^2$

Initial velocity=u=$\sqrt{\frac{2E}{m}}$

Final velocity=0

Friction force=$\mu mg$

Where g=Acceleration due to gravity

Acceleration due to friction force=$-\frac{\mu mg}{m}=-\mu g$

(negative sign shows retardation)

$v^2-u^2=2as$

$0-\frac{2E}{m}=2(-\mu g) S$

$S=\frac{2E}{m\times \mu g}=\frac{2E}{\mu mg}$

Hence, the stopping distance=$\frac{2E}{\mu mg}$