AP EAPCET 07-Jul-2022 Shift 2 Paper

Given, displacement of block, x=5t2
As, velocity v=

dx

dt

Velocity of block, v=

d

dt

(5t2)
=5×2t=10tm∕ s
Now, velocity of block at t=5 s,
v2=10×5=50m∕ s
and velocity of block at t=3 s is
v1=10×3=30m∕ s
Also, velocity of block at t=0 s,
v0=10×0=0m∕ s
Now, by work- K.E theorem, work done by a force = Change in kinetic energy
So work done by force in first 3 sec,
W1=

1

2

mv12−

1

2

mv02=

1

2

m(v12−v02)
And work done by force in first 5 sec,
W2=

1

2

mv22−

1

2

mv02=

1

2

m(v22−v02)
Hence, ratio

W1

W2

=

1 2 m(v12−v02)

1 2 m(v22−v02)

=

v12−v02

v22−v02

or

W1

W2

=

v12

v22

...[As,v0=0]
⇒

W1

W2

=(

30

50

)2=

9

25