KVPY SA Exam 2014 Question Paper

When box is dropped from a height h, then speed at ground is v, therefore using mechanical energy conservation
mgh=

1

2

mv2 …(i)
when body slides on rough inclined plane, friction force will also act f = μN = μ mg cos θ
Applying work-energy theorem
mgh‌–‌fs=

1

2

m(

v

3

)2−0
mgh‌–‌f‌

h

sin‌θ

=

1

2

m(

v

3

)2
(sin‌θ=

h

s

)
mgh−μ‌mg‌cos‌θ×

h

sin‌θ

=

1

2

mv2

9

…(ii)
from equation (i) & (ii)
mgh[1−μ‌cot‌θ]=(

1

9

)‌mgh
putting θ = 45º, cotθ = 1
1−μ=

1

9

⇒ μ=

8

9