force friction= .3*mg
Vf^2=Vi^2+2ad but a= -forcefricion/m
Vf=0, Vi is known, solve for d.
Physics force of friction question
a shuffle board disk is accelerated to a speed of 5.8m/s and released. if the coefficient of the kinetic friction between the disk and the concrete court is 0.3,
how far does the disk go before it comes to a stop?The courts are 15.8meters long
4 answers
isn't vf 5.8
F=m*a So u*m*g=m*a
m cancels out cause its on both sides of the equation.
so u*g=a (mu*gravity=acceleration)
solve for acceleration then plug a into the formula vf^2=Vi^2+2*a*d to solve for d.
m cancels out cause its on both sides of the equation.
so u*g=a (mu*gravity=acceleration)
solve for acceleration then plug a into the formula vf^2=Vi^2+2*a*d to solve for d.
Wrong. 9+10=21
1 answer