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