Question

A solid, spherical ball slides with speed 4.48 m/s across frictionless ice and then encounters a rough patch with a coefficient of kinetic friction µk = 0.198. The ball is initially not rotating but the friction causes it to start spinning. How far does it slide across the rough spot before it begins to roll without slipping? Express your answer in metres.

Answer 1

acceleration acting a = -uk*g

after travelling a distance d

v2^2 - v1^2 = 2*a*d

v2^2 - v1^2 = -2*uk*g*d

v2^2 = v1^2 - 2*uk*g*d

kinetic energy after travelleing distance

K2 = (1/2)*m*v2^2 + (1/2)*I*w^2

K2 = (1/2)*m*v2^2 + (1/2)*(2/5)*m*r^2*(v2/r)^2

K2 = (7/10)*m*v2^2

work done by friction

Wf = -uk*m*g*d

from work energy relation

work = change in KE

W = K2 - K2

-uk*m*g*d = (7/10)*m*v2^2 - (1/2)*m*v1^2

- uk*g*d = (7/10)*(v1^2 - 2*uk*g*d) - (1/2)*v1^2

-0.198*9.81*d = (7/10)*(4.48^2-(2*0.198*9.81*d)) - (1/2)*4.48^2

distance d = 5.17 m <<<------answer