If a #3kg# object moving at #1 m/s# slows down to a halt after moving 2 m, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Andrea B.
Jun 21, 2018

#mu= 0,0255#

Explanation:

initial kinetic energy
#E_k = 1/2 m xx v^2 = 1/2 xx 3 kg xx 1 m^2/s^2 = 1,5 J #
if the object will stop after 2 m this energy is lost with the friction work
# W_f = F xx s = mu xx g xx m xx s= mu xx 9,8 m/s^2 xx 3 kg xx 2 m = mu xx 58,8J#
since
#E_k= W_f#
#1,5 J = mu xx 58,8 J#
friction coefficient
#mu = 0,0255#

by the second dynamic principle
# F= m xx a #
#a= F/m = (mu xx g xx m)/ m = mu xx g = 0,25 m/s^2#
by the cinematic equation
# v= -a xx t + v°#
#t= v°/a = (1 m/s)/(0,25 m/s^2)= 4 s#
#s= -1/2 a xx t^2+ v° xx t= 1/2 xx 0,25 m/s^2 xx (4s)^2 + 1m/s xx 4 s=-2m + 4 m = 2m#
and the datas are verificated

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