If an object with a mass of #10 kg # is moving on a surface at #45 m/s# and slows to a halt after # 6 s#, what is the friction coefficient of the surface?

1 Answer
Mar 28, 2016

Answer:

#u_k=0,0051#

Explanation:

#v_l=v_i-a*t#
#v_l:"last velocity"#
#v_i:"initial velocity"#
#a:"acceleration"#
#t:"time"#

#v=v/2-a*6#
#6*a=v-v/2#
#6*a=-v/2#
#a=-v/12" "a=-3,75" " m/s^2" 'acceleration of object'"#

#v_l^2=v_i^2-2*a* Delta x#
#v^2=(v/2)^2-2.3,75*Delta x#

#v^2-v^2/4=7,50*Delta x#

#(3*v^2)/4=7,50*Delta x#
#3*v^2=30*Delta x#
#v^2=10*Delta x" ; "45^2=10*Delta x" ; "2025=10*Delta x#
#Delta x=202,5 " m"#
#E_k=1/2*m*v^2 "the kinetic energy of object"#
#W_f=F_f*Delta x" work doing by friction force"#
#F_f=u_k*m*g" "W_f=u_k*m*g*Delta x#
#E_k=W_f#
#1/2*cancel(m)*v^2=u_k*cancel(m)*g*Delta x#

#u_k=v^2/(2*g*Delta x)#

#u_k=45^2/(2*9,81*202,5)=2025/397305#

#u_k=0,0051#