A .5 kg block is sliding along a tabletop with an initial velocity of .20 m/s. It slides to rest in a distance of .7m. Find the frictional force regarding it motion.
The answer given is .0143 N but I do not understand how to get that.
2 answers
2.3m/s 17m/s
Frictional Force=1/2mv^2/d
=1/2(0.5)(0.2)^2/0.7
=1/2(0.5)(0.04)/0.7
=1/2(0.02)/0.7
=0.01/0.7
=0.0142857143
=0.0143 N
=1/2(0.5)(0.2)^2/0.7
=1/2(0.5)(0.04)/0.7
=1/2(0.02)/0.7
=0.01/0.7
=0.0142857143
=0.0143 N