If a #2 kg# object moving at #5 m/s# slows down to a halt after moving #3 m#, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Feb 21, 2016

Answer:

#u_k~=0,425#

Explanation:

#E_k=1/2*m*v^2 " The kinetic energy of object"#
#W=F_f* Delta x " Work doing by friction force"#
#F_f=u_k*N " N: normal force to contacting surfaces"#
#"The kinetic energy turns work"#
#1/2*cancel(m)*v^2=u_k.cancel(m)*g*Delta x#
#1/2*5^2=u_k*9,81*3#
#25=u_k*2*3*9,81#
#u_k=25/(2.3.9,81)#
#u_k~=0,425#