2. A donkey pulls a 100kg sled for 20m across a level surface.
a. If the rope pulls at a 30° angle to the horizontal with just enought energy
to force to move the sled without acceleration and the coefficient of friction
for the sled/surface is 0.3, how much work does the donkey do?
b. If the donkey does all this in 10 seconds, how much power is it exerting?
a. 5787.402J
b. 578.7402W
Where as in most problems, we find the components of the force from the force itself, in this
instance we need to find the force of the donkey (FD) from the components. We are going to make some equations
using the information we have and then solve for the force. Once that is done, all we have to do
is multiply that by 20m.
FG = 100kg * 9.8m/s^2
FG = 980N
980N = FN + FD sin 30°
FN = 980N - FD sin 30°
FD cos 30° = FK
FK = 0.3 * FN
FD cos 30° = 0.3 * FN
FD cos 30° = 0.3 * (980N - FD sin 30°)
FD * 0.866 = 294N - FD * 0.15
FD * 1.016 = 294N
FD = 289.37N
W = 289.37N * 20m
W = 5787.402J
In part b, we just use the equation for power and divide the work by the time.
P = 5787.402J / 10s
P = 578.7402W
