A cylinder of mass 12 ~kg is sliding on plane with an initial velocity 20 ~ms ⁻¹ . If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping, the cylinder describes a distance of

Step-by-step solution

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