A 0.5 kg block is sliding along a table top with an initial velocity of 0.2 m/s. It slides to rest in a distance of 70 cm. Find the frictional force that is slowing its motion.

Answer 1

KE=W(fr)
KE = m•v²/2
W(fr) =F(fr) •s
m•v²/2 =F(fr) •s
F(fr)= m•v²/2•s

Answer 2

142.85