AP EAPCET 08-Jul-2022 Shift 1 Paper

Given, Force on block, F=F

^

i

Acceleration of block, a=β.x=5x
Now, let instantaneous velocity of block is v, then
dv∕dt=a

⇒⇒

dv

dx

dx

dt

=a
⇒v

dv

dx

=a
or vdv=a‌dx
Now, integrating between limits x1=2×10−2m to
x2=5×10−2m
⇒

∫v2

v1

v⋅dv=

‌

∫

x1

x25x⋅dx
⇒[

v2

2

]v1v2=

5

2

[x2]x1x2
⇒

1

2

(v22−v12)=

5

2

(x22−x12)
Here,
x1=2‌cm and x2=5‌cm
so
⇒

1

2

(v22−v12)=

5

2

(52−22)×(10−2)2
Multiplying with mass of body m=1‌kg , we get

1

2

m(v22−v12)=

5

2

×m×21×10−2
⇒ Change in kinetic energy
=

5

2

×1×21×(10−2)2=525×10−4
As change in KE=Work‌done
∴Work‌done=52.5×10−4J