Question

A body of mass 1 kg starts moving from rest at t= 0, in a circular path of radius 8 m. Its kinetic energy varies
as a function of time as : K.E. = 2t2 Joules, where t is in seconds. Then
A tangential acceleration = 4 m/s2
(B) power of all forces at t = 2 sec is 8 watt
(C) first round is completed in 2 sec.
(D) tangential force at t = 2 sec is 4 newton.​

Answer 1

Given:

A body of mass 1 kg starts moving from rest at t= 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as : K.E. = 2t² J.

To find:

Correct Option among the above ?

Calculation:

$\therefore \: KE = 2 {t}^{2}$

$\implies \: \dfrac{1}{2} m {v}^{2} = 2 {t}^{2}$

$\implies \: \dfrac{1}{2} (1) {v}^{2} = 2 {t}^{2}$

$\implies \: {v}^{2} = 4{t}^{2}$

$\implies \:v = 2t$

$\implies \:a_{t}= \dfrac{dv}{dt}$

$\implies \:a_{t}= \dfrac{d(2t)}{dt}$

$\implies \:a_{t}= 2 \: m {s}^{ - 2}$

So, tangential acceleration will always be 2 m/s².

$\implies \:v = 2t$

$\implies \: \dfrac{dx}{dt} = 2t$

$\implies \displaystyle \int_{0}^{2\pi(8)} \:dx = \int_{0}^{t} 2t \: dt$

$\implies \: 2\pi (8) = {t}^{2}$

$\implies \: {t}^{2} = 16\pi$

$\implies \: t= 4 \sqrt{\pi} \: sec$

So, first round is completed in 4√π sec.

Answer 2

Explanation:

2

1

mv

2

=2t

2

v

2

=4t

2

⇒v=2t

a

t

=

dt

dv

⇒a

t

=2m s

−2

F

t

=ma

t

⇒F

t

=2N

Power will be inverted only by tangential force

P=(

F

t

+

F

c

)⋅

V

since F

c

⊥V&F

t

∥V

P=F

t

V⇒P=2×2t

⇒P=4t⇒P=8W