Question

Open in App

Solution

Verified by Toppr

Correct option is C)

Case $I$ :

$[2u ]_{2}−u_{2}=2.a.3$or $−43u_{2} =2.a.3⇒a=−8u_{2} $

Case $II$ :

$0−[2u ]_{2}=2.a.x$ or $−4u_{2} =2[−8u_{2} ]×x$

$⇒x=1cm$

Alternative method : Let K be the initial energy and F be the resistive force. Then according to work-energy theorem,

$W=ΔK$

i.e., $3F=21 mv_{2}−21 m[2v ]_{2}$

$3F=21 mv_{2}[1−41 ]$

$3F=43 [21 mv_{2}]$ ............(1)

and $Fx=21 m[2v ]_{2}−21 m(0)_{2}$

i.e., $41 [21 mv_{2}]=Fx$ .........(2)

Comparing eqns. (1) and (2)

$F=Fx$

or $x=1cm$

Solve any question of Work, Energy and Power with:-

Was this answer helpful?

0

0