If an object is moving at #50 m/s# over a surface with a kinetic friction coefficient of #u_k=8 /g#, how far will the object continue to move?

1 Answer
Dec 11, 2015

Answer:

I found #156m#

Explanation:

Considering that the only force acting on the object will be kinetic friction #f_k#, it will produce an opposite acceleration slowing down the object.
From Newton's Second Law (horizontally):

#-f_k=ma#

and so:
#a=-f_k/m#

Friction will be: #f_k=u_kN#
where, in this case, the normal reaction, #N#, will be equal to the weight #W=mg#;
so:
#a=-(u_kcancel(m)g)/cancel(m)=-8/g*g=-8m/s^2#
With this acceleration you can use:
#color(red)(v_f^2=v_i^2+2ad)#
with #v_f=0# (the object stops)
so you get:
#0^2=50^2-(2*8*d)#
#d=2500/(16)=156m#