AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The figure below shows the free body diagram of the block.The friction force acting on the block is given as,
f=µk(mg‌cos‌θ)
The force equation for the block along the plane of inclination,
mg‌sin‌θ−f‌‌=ma
mg‌sin‌θ−µk(mg‌cos‌θ)‌‌=ma
a‌‌=10×0.6−‌

1

8

×10×0.8
a‌‌=5m∕s2
The distance s is given from ∆EBC
EC=s=‌

BE

sin‌θ

=‌

25 6

0.6

=‌

125

18

m
The velocity at final position is give as,
v2=2as
Substituting the values in above expression, we get
v2=2×5m∕s2×‌

125

18

m
v=‌

25

3

m∕s
The component of velocity at point C is given as,Using second equation of motion, we get
h‌‌=ut+‌

1

2

gt2
10‌‌=ut‌sin‌θ+‌

1

2

gt2
5t2+5t−10‌‌=0
t2+t−2‌‌=0
On solving the time comes out to be t=1 s The distance DF is,
DF=v‌cos‌θ=‌

25

3

×0.8=‌

20

3

m