A block of mass m slides down an incline of mass M. As shown, th acceleration of the block is a. The coefficient of kinetic friction between the block and the incline is ì1, while the coefficient of kinetic friction between the incline and ground is ì2. If the incline stays motionless all the time, what is the friction f between incline and ground, and what is the normal force N vertically acting on the ground?
Select one:
a. f = (M+m)gì2 N = (M + m)g - ma sinè
b. f = mg(cos è - sin è)cosè N = (M + m)g - ma cosè
c. f = macos è N =Mg + mg cos è (cos è + ì1sin è)
d. f = mg ì1cos è N= (M + m)g
e. f = Mgì2 N = Mg
To solve this problem, we can break it down into two separate components: the motion of the block sliding down the incline and the forces acting on the incline itself.
1. Motion of the block sliding down the incline:
The force of gravity acting on the block can be resolved into two components: a component parallel to the incline (m * g * sinθ) and a component perpendicular to the incline (m * g * cosθ). The friction force opposing the motion of the block on the incline can be calculated using the equation f = μ1 * N, where μ1 is the coefficient of kinetic friction between the block and the incline, and N is the normal force acting on the block. The net force acting on the block in the parallel direction is given by F_net = m * a, where a is the acceleration of the block.
Using Newton's second law, we can write the equation for the motion of the block in the parallel direction as:
m * a = m * g * sinθ - μ1 * N
Note that since the incline stays motionless, the normal force acting on the incline will cancel out the normal force acting on the block, leaving us with only the normal force acting on the ground.
2. Forces acting on the incline:
The incline is in equilibrium, so the net force acting on it in the horizontal direction is zero. The only force acting in the horizontal direction is the friction force between the incline and the ground. Therefore, we can write the equation for the horizontal direction as:
f = μ2 * N.
Now, let's analyze the answer choices:
a. f = (M+m)gμ2 N = (M + m)g - ma sinθ
This answer choice correctly represents the equation for the friction force between the incline and the ground (f = μ2 * N), but the equation for the normal force vertically acting on the ground is incorrect.
b. f = mg(cosθ - sinθ)cosθ N = (M + m)g - ma cosθ
This answer choice does not reflect the correct equation for the friction force (f = μ2 * N). The equation for the normal force is also incorrect.
c. f = ma * cosθ N = Mg + mg cosθ (cosθ + μ1sinθ)
This answer choice does not represent the equation for the friction force between the incline and the ground (f = μ2 * N). The equation for the normal force is also incorrect.
d. f = mgμ1cosθ N = (M + m)g
This answer choice correctly represents the equation for the friction force between the incline and the ground (f = μ2 * N), and the equation for the normal force vertically acting on the ground N = (M + m)g.
e. f = Mgμ2 N = Mg
This answer choice correctly represents the equation for the friction force between the incline and the ground (f = μ2 * N), but it does not reflect the correct equation for the normal force vertically acting on the ground.
Therefore, the correct answer is: d. f = mgμ1cosθ, N = (M + m)g.