posted by james .
Masses 19 kg and 9 kg are connected by a light string that passes over a frictionless pulley as shown in the figure ( The figure shows the 19 kg block on the table connected by a string through a pulley which is connected to the 5 kg block hanging off the table). The acceleration due to gravity is 9.8m/s. If the 19 kg mass, initially held at rest on the table, is released and moves 1.2 m in 1.4s, determine the coefficient of kinetic friction between it and the table.
Ok, work done by gravity on the 9 kg block is put into friction, and the final KE of both blocks. One then has to find the final velocity of the blocks.
Energy done by gravity= 9g*1.2
energy going into friction= mu*19g*1.2
energy going into ke= 1/2 (27)vfinal^2
distancewent=1/2 a t^2 solve for a
1.2=1/2 a (1.4^2) solve for a.
then vfinal= at= a*1.4
vfinal=1.2*2/1.96 *1.4 check that
vfinal^2= 2.94 check that.
so you have the pieces.
9g*1.2=19g*mu*1.2+1/2 (27)(2.94 )
and you solve for mu