A 50 kg trunk is pushed 6.6 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is 0.22. a)Calculate the work done by the applied horizontal force. b)suppose the 50 kg trunk is pushed 6.6 m at constant speed up a 30° incline by a force along the plane. The coefficient of kinetic friction between the trunk and the incline is 0.22. Calculate the work done by the applied force. c)How much energy was dissipated by the frictional force acting on the trunk?
2007-11-09 13:30:40 · 1 answers · asked by wonderwoman 1 in Science & Mathematics ➔ Physics
The component of horizontal force Fh is doing all the work against downward force Fd caused by weight of the trunk and the friction force f. We can express it as
Fh=(Fd + f )/cos(30) however at the moment we are after work
a) W=(Fd + f) s
Fd= mg sin(30)
f=umg cos(30)
s= 6.6 m
now
W= mg( sin(30) + u cos(30)) s
W= 50 x 9.81 x(sin(30) + .22 cos(30)) 6.6=
W=2240 J
b) same as a) since we are always interested only in the force along the line of motion.
c) Wf= u mgcos(30) s
wf= .22 x 50 x 9.81x cos(30)) x 6.6=617 J
2007-11-09 13:35:03 · answer #1 · answered by Edward 7 · 0⤊ 3⤋