A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m / s at t=0 s on a rough horizontal surface. It starts off with a purely sliding motion at t=0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be J (given, coefficient of friction is 0.3 and .g =10 m / s 2).

Question

A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m / s at t=0 s on a rough horizontal surface. It starts off with a purely sliding motion at t=0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be J (given, coefficient of friction is 0.3 and .g =10 m / s 2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a =-μ k g =-3 V =18-3 × 2 V =12 m / s KE =(1/2) mv 2+(1/2) ( mr 2/2) ( v 2/ r 2) KE =(3/4) mv 2 KE =3 × 18=54 J

a=−μk​g=−3 V=18−3×2 V=12m/s KE=21​mv2+21​2mr2​r2v2​ KE=43​mv2 KE=3×18=54J

$a = - μ_{k} g = - 3$
$V = 18 - 3 \times 2$
$V = 12 m / s$
$K E = \frac{1}{2} m v^{2} + \frac{1}{2} \frac{m r ^{2}}{2} \frac{v ^{2}}{r ^{2}}$
$K E = \frac{3}{4} m v^{2}$
$K E = 3 \times 18 = 54 J$