If a #13 kg# object moving at #400 m/s# slows down to a halt after moving #20 m#, what is the friction coefficient of the surface that the object was moving over?

Question

2269 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T12:42:35

The coefficient of friction is #=408.2#

Apply the equation of motion to find the acceleration

#v^2=u^2+2as#

The initial velocity is #u=400ms^-1#

The final velocity is #v=0ms^-1#

The distance is #s=20m#

The acceleration is

#a=(v^2-u^2)/(2s)=(0-400^2)/(2*20)=-4000ms^-2#

The mass of the object is #m=13kg#

According to Newton's Second Law of Motion

#F=ma#

The force of friction is #F_r=13*4000=52000N#

The acceleration due to gravity is #g=9.8ms^-2#

The normal reaction is #N=mg=13*9.8=127.4N#

The coefficient of kinetic friction is

#mu_k=F_r/N=52000/127.4=408.2#