If an object with a mass of 10 kg 10kg is moving on a surface at 45 m/s45ms and slows to a halt after 6 s6s, what is the friction coefficient of the surface?

1 Answer
Mar 28, 2016

u_k=0,0051uk=0,0051

Explanation:

v_l=v_i-a*tvl=viat
v_l:"last velocity"vl:last velocity
v_i:"initial velocity"vi:initial velocity
a:"acceleration"a:acceleration
t:"time"t:time

v=v/2-a*6v=v2a6
6*a=v-v/26a=vv2
6*a=-v/26a=v2
a=-v/12" "a=-3,75" " m/s^2" 'acceleration of object'"a=v12 a=3,75 ms2 '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