a 3 kg block slides down a frictionless plane inclined 20 degrees to the horizontal. If the length of the plane's surface is 1.50m, how much work is done, and by what force?
suppose the coefficient of kinetic friction between the block and the plane. What would be the net work done in this case?
3 answers
sorry i forgot to add the coefficient of friction is .275
Gravity does the work which is equal to the change in potential energy, m g h
here h = 1.50 sin 20
so
m g h = 3*9.8 *1.5*sin 20
friction will do negative work
mu m g cos 20 * 1.50
subtract that from m g h to get net
here h = 1.50 sin 20
so
m g h = 3*9.8 *1.5*sin 20
friction will do negative work
mu m g cos 20 * 1.50
subtract that from m g h to get net
So mu = .275
1 answer