TS EAMCET 04 May 2019 Shift 1 Solved Paper

Consider the below diagram
The equation of acceleration in downwards motion of the block is written as,
a=g‌sin‌θ−µg‌cos‌θ
Substitute the given values as
a‌‌=g(sin‌45∘−0.3‌cos‌45∘)
‌‌=‌

10

√2

(1−0.3)
‌‌=‌

10×0.7

√2

m

s2

The displacement of the block is calculated from the second equation of motion as,
s‌‌=ut+‌

1

2

at2
‌‌=‌

1

2

×‌

10×0.7

√2

×(2)2
‌‌=‌

7×4

2√2

‌‌=7√2m