You push a 34.4 kg crate at a constant speed of 1.66 m/s across a horizontal floor with coefficient of kinetic friction 0.34. The acceleration of gravity is 9.8 m/s^2.
a) At what rate is work being done on the crate by you? Answer in W.
b) At what rate is energy being dissipated by the frictional force? Answer in W.
2007-10-01 17:19:44 · 1 answers · asked by Ron108 1 in Science & Mathematics ➔ Physics
The normal force of the box is
Fn = 34.4*9.8 = 337.12 N so the frictional force is
Ff = 337.12*.34 = 114.62 N Since the crate is not accelerating, that means the you are pushing with a force of 114.62N and the work done is of course force times distance. Since the crate is moving 1.66 meters/sec, then in one second you have done
114.62*1.66 = 190.23 J of work. And, since you did it in 1 second and power is J/s, you're power output is 190.23 Watts.
Doug
2007-10-01 17:32:31 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋