TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, angle of inclination, θ=30∘.
Coefficient of friction µ=‌

2

3

x, (where x is distance)
and final speed, v=0 From given diagram,
Net force, ma=mg‌sin‌θ−µmg‌cos‌θ
a=g(sin‌θ−µ‌cos‌θ)
where, a is acceleration.
∴‌‌a‌‌=10(sin‌30∘−‌

2

3

x‌cos‌30∘)
a‌‌=10(‌

1

2

−‌

2

3

x⋅‌

√3

2

)
‌‌=10(‌

1

2

−‌

1

√3

x)
We know that,
a‌‌=v‌

dv

dx

⇒‌‌vdv‌‌=adx . . . (i)
where, v is velocity and dv,dx are elemental velocity and displacement, respectively.
Now, on integrating Eq. (i) both sides, we get
‌

v=0

∫

u=0

vdv‌‌=

x

∫

0

10(‌

1

2

−‌

1

√3

x)dx
‌∴‌(‌

v2

2

)00‌‌=10(‌

x

2

−‌

1

√3

‌

x2

2

)0x
⇒‌0‌‌=‌

10

2

(x−‌

x2

√3

)
⇒‌x‌‌=‌

x2

√3

⇒‌x‌‌=√3m