A piece of stone came to rest after covering a distanceof 20 m over an icy surface. If the initial velocity of thestone was lm.s-1, what was the coefficient of frictionbetween the stone and ice?[0.0026]​

Answer:

A piece of stone, moving with an initial VELOCITY of 11 m/s on an icy surface, comes to rest after covering a DISTANCE of 2020 m. We want to determine the coefficient of friction between the stone and ice.

The initial velocity, uu, final velocity, vv, and distance covered, s,s, are 11 m/s, 00 m/s and 2020 m, respectively.

v2=u2+2as,v2=u2+2as, where AA is the acceleration.

⇒02=12+2a×20⇒a=−140⇒02=12+2a×20⇒a=−140 m/s2.2.

⇒⇒ The magnitude of the force on the stone causing it to slow down and finally come to a stop is ma=m40ma=m40 N, where mm is the mass of the stone.

The force slowing down the stone is due to friction and is equal to μN,μN, where μμ and NN are the coefficient of KINETIC friction and the normal reaction, respectively.

The normal reaction, N=mg,N=mg, where gg is the acceleration due to gravity, which can be taken as 9.819.81 m/s22.

⇒μmg=|ma|⇒μ=