1a) The equation you will need to solve this problem is F = ma.
The "F" part of the equation is the net force, which is the sum of the force applied by the donkey (F1), the weight component up the hill (F2) and the friction force (F3).
F = F1 - F2 - F3 = m a
= 100 kg*0.2 m/s^2 = 20 N
F2 = m g sin30 = 980 * (0.5) = 490 N
The friction force is
F3 = m g cos 30*Uk = 255 N
where Uk is the kinetic friction coefficient. 0.3 in this case.
Solve for F1.
1b) Think: Newton's Third law.
1c) mg cos 30
1d) M is now the donkey's mass. It applies both a normal force (M g sin 30) and a friction force (M a - F1) to the incline. They don't tell you the value of M
1e) a = [m g sin 30 - m g cos 30 Uk]/m
= g(0.5 - 0.26) = ___
1a)A block of mass=100kg is tied to a donkey that moves with constant acceleration of 20cm/s^2 up a 30 degree incline. What is the force with which the donkey pulls on the block parallel to the incline? the coefficient of kinetic friction between the incline and the block is 0.30.
1b) how large is the force with which the block pulls on the donkey?
1c) how large is the normal force?
1d) what is the force that the donkey of mass M,applies on the incline?
1e) if the rope that ties the box to the donkey is cut, what will the acceleration of the box be?
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