At a certain factory, 300 kg crates are dropped vertically from a packing machine onto a conveyor belt moving at 1.20 m/s. The coefficient of kinetic friction between the belt and each crate is 0.400. After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate and (c) the energy supplied by the motor. (d) explain why answers (a) and (c) differ.
2006-10-27 08:10:23 · 2 answers · asked by afchica101 1 in Science & Mathematics ➔ Physics
a. KE=1/2mv^2
=1/2*300*1.20^2
=216 Joules
b. Frictional force =uR where u is the coefficient of kinetic friction, and R is the weight of the crate pressing on the belt=300*g.
Frictional force=0.400*300*9.8
=1172 N
c. Same as in a=216 Joules, unless you include the additional energy required to overcome friction within the belt and the rollers underneath it. But that is totally an unknown quantity.
2006-10-28 04:53:30 · answer #1 · answered by tul b 3 · 0⤊ 0⤋
do your own homework!
2006-10-27 08:19:00 · answer #2 · answered by bikeworks 7 · 0⤊ 3⤋