Question

Starting from rest, a body slides down a 45o inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is  ​

Answer 1

Answer:

Case (1) When there is no friction,

From newton's second law, F=ma

⇒mgsinθ=ma

1

⇒a

1

=gsinθ

Now, as body starts from rest so initial velocity (u)=0 m/s.

Let, body travels 'x' distance in t

1

time,

So from equation of motion,

x=ut+

2

1

at

2

⇒x=

2

1

a

1

t

1

2

⇒t

1

=

a

1

2x

=

gsinθ

2x

Case (2) When there is friction,

From newton's second law, F=ma

⇒mgsinθ−μmgcosθ=ma

2

⇒a

2

=gsinθ−μgcosθ

Now, as body starts from rest so initial velocity (u)=0 m/s.

Let, body travels 'x' distance in t

2

time,

So from equation of motion,

x=ut+

2

1

at

2

⇒x=

2

1

a

2

t

2

2

⇒t

2

=

a

1

2x

=

gsinθ−μgcosθ

2x

It is given that t

2

=2t

1

So,

gsinθ−μgcosθ

2x

=2

gsinθ

2x

Now, substituting θ=45 and solving above equation we get,

1−μ

1

=2

μ=

4

3

=0.75

Answer 2

Answer:

I don't know ji

but app

mere questions ke answer bhi dedo ji