AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

The free body diagram of the system is sown below,The acceleration of the block is,
a=µg
a=(0.2)(10)
a=2m∕s2
The initial velocity of the block is calculated as
v2=u2−2as
(0)2‌‌=u2−2(2)(1)
u‌‌=2m∕s
The net accelerating force down the inclined plane is,
F‌‌=ma1
mg‌sin‌30∘−µmg‌cos‌30∘‌‌=ma1
a1‌‌=g(sin‌30∘−µ‌cos‌30∘)
a1‌‌=5(1−√3µ)
The displacement is calculated as
v12‌‌=u12+2a1S
(2)2‌‌=(0)2+2[5(1−√3µ)]S‌( As, ‌v1=u=2m∕s,u1=0)
S‌‌=‌

2

5(1−√3µ)

S‌‌=0.612m
The work done by friction force is,
W=(µmg‌cos‌30∘)S+‌

1

2

mu2
W=(0.2)(‌

100

1000

)(10)(cos‌30∘)(0.612)+‌

1

2

(‌

100

1000

)(2)2
W=0.306J