Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N; the component of the force from the donkey in the direction of motion isF2 = F1
[email protected]= 397.2964498768165 N
Fy = N - mg
[email protected] = 0N = mg
[email protected] = 4037.964151113007 NFx = F2 - mg
[email protected] - f = mahere f = u N=65.8188156631420141
F2 - mg
[email protected] - f = maa = F2 - mg
[email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
Answer:
C. An inital volocity that is faster than the final volocity
Explanation:
.
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
