AP EAPCET 23-Aug-2021 Shift 2 Paper

Given that, mass of cylinder, m=12‌kg
Initial velocity, u=20m∕s
Coefficient of friction, µ=0.5
We know that, the retardation produced by friction =a=−µg =−0.5×10 =−5m∕s2
[using, g=10m∕s2]
Let S be the distance travelled by cylinder to stop.
[i.e. final velocity, v=0 ]
Using equation of motion, v2=u2+2as
By substituting the value, we get
0=(20)2+2(−5)s
⇒s=40m