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?

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) = ___

To solve these questions, we will use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. We will also consider the forces acting on the block on the incline: the gravitational force, the force of friction, and the force exerted by the donkey.

1a) The force with which the donkey pulls the block parallel to the incline can be calculated by considering the forces acting on the block. We need to find the force parallel to the incline. The perpendicular component of gravity will be balanced by the normal force, so it does not contribute to this force. The only force in the parallel direction is the force of friction. The formula for frictional force is F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force can be calculated by resolving the gravitational force into its components, Fs = m * g * sin(θ) and perpendicular component Fp = m * g * cos(θ), where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline. The normal force N = Fp = m * g * cos(θ). Thus, the force with which the donkey pulls on the block parallel to the incline is F_parallel = μ * N.

1b) The force with which the block pulls on the donkey can be determined using Newton's third law of motion, which states that every action has an equal and opposite reaction. Therefore, the force exerted by the block on the donkey is the same as the force exerted by the donkey on the block, which we calculated in 1a).

1c) The normal force can be calculated using the formula N = m * g * cos(θ), where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline.

1d) To find the force that the donkey applies on the incline, we consider the forces acting on the donkey. The only external force acting on the donkey parallel to the incline is the force exerted by the block, which we calculated in 1b).

1e) If the rope is cut, the block will be free from the force exerted by the donkey. The only force acting on the block will then be the component of gravitational force parallel to the incline, which is Fs = m * g * sin(θ). To find the acceleration of the block, we can use Newton's second law of motion, F = m * a, where F is the net force acting on the block (in this case, the force due to gravity parallel to the incline), m is the mass of the block, and a is the acceleration. Rearranging the formula, we have a = F / m. By substituting Fs for F, we get a = (m * g * sin(θ)) / m, which simplifies to a = g * sin(θ). Therefore, the acceleration of the block will be g * sin(θ).