At the point where the block is about to slide, the static friction force has reached its maximum value, $$F_{s,max} = \mu_s F_N$$. This force must balance the component of gravity parallel to the incline, $$mg \sin\theta_{max}$$. The normal force is $$F_N = mg \cos\theta_{max}$$. Setting the forces parallel to the incline equal: $$mg \sin\theta_{max} = \mu_s (mg \cos\theta_{max})$$. The term $$mg$$ cancels from both sides, leaving $$\sin\theta_{max} = \mu_s \cos\theta_{max}$$. Solving for $$\mu_s$$ gives $$\mu_s = \sin\theta_{max} / \cos\theta_{max} = \tan\theta_{max}$$.

If the block remains at rest, it is in static equilibrium, and the net force on it is zero. The applied horizontal force $$F$$ is balanced by the static friction force $$f_s$$. Therefore, the magnitude of the static friction force must be equal to the magnitude of the applied force, $$f_s = F$$. This is true as long as $$F$$ does not exceed the maximum static friction, $$F \le \mu_s Mg$$. The static friction force adjusts its magnitude to match the applied force.

The block will begin to move when the applied force $$F(t)$$ becomes equal to the maximum static friction force, $$F_{s,max} = \mu_s F_N$$. For a block on a horizontal surface, $$F_N = mg$$. So, the block starts to move when $$F(t) = \mu_s mg$$. Since $$F(t) = ct$$, we set $$ct = \mu_s mg$$. Solving for $$t$$ gives $$t = \mu_s mg / c$$.

Since the block is at rest, it is in static equilibrium. The net force on the block is zero. The forces acting parallel to the incline are the component of gravity pulling the block down the incline ($$mg \sin\theta$$) and the static friction force ($$F_s$$) acting up the incline. Therefore, $$F_s = mg \sin\theta$$. Using $$g \approx 9.8 , \text{m/s}^2$$, $$F_s = (5.0 , \text{kg})(9.8 , \text{m/s}^2)(\sin 30°) = (49 , \text{N})(0.5) = 24.5 , \text{N}$$. The static friction force adjusts its magnitude to be equal and opposite to the net force that would cause motion, as long as that force does not exceed the maximum static friction.

First, apply Newton's second law in the horizontal direction: $$F_{net} = F_{applied} - F_k = ma$$. The kinetic friction force is $$F_k = F_{applied} - ma = 200 , \text{N} - (50 , \text{kg})(1.0 , \text{m/s}^2) = 150 , \text{N}$$. The magnitude of the kinetic friction force is also given by $$F_k = \mu_k F_N$$. On a level floor, $$F_N = mg = (50 , \text{kg})(9.8 , \text{m/s}^2) = 490 , \text{N}$$. Therefore, $$\mu_k = F_k / F_N = 150 , \text{N} / 490 , \text{N} \approx 0.306$$, which is closest to 0.31.

The block begins to move when the applied horizontal force exceeds the maximum force of static friction, $$ F_{s,max} = \mu_s F_N $$. On a horizontal surface, the normal force $$ F_N $$ is equal to the weight of the block, $$ mg $$. Using $$ g \approx 9.8 , \text{m/s}^2 $$, $$ F_N = (10 , \text{kg})(9.8 , \text{m/s}^2) = 98 , \text{N} $$. Therefore, $$ F_{s,max} = (0.50)(98 , \text{N}) = 49.0 , \text{N} $$. The block will start to move when the applied force is just greater than 49.0 N.

This question tests AP Physics C: Mechanics skills, specifically understanding of kinetic and static friction and their roles in force dynamics. Kinetic friction acts on moving objects with a constant magnitude equal to μk times the normal force, independent of the object's speed or applied force. In this scenario, a 10.0 kg sled is already moving on packed snow with μk = 0.10, and a student pulls with 30 N horizontally. Choice A (9.8 N) is correct because kinetic friction equals μk × N = μk × mg = 0.10 × 10.0 × 9.8 = 9.8 N, regardless of the 30 N applied force. Choice B (19.6 N) incorrectly uses the static coefficient or doubles the kinetic friction, while choice C uses the applied force magnitude. Students should understand that kinetic friction depends only on the coefficient and normal force, not on the applied force or velocity. Draw free-body diagrams showing all forces and emphasize that kinetic friction opposes motion with constant magnitude.

The kinetic friction force always opposes the direction of relative motion. Since the box is sliding to the right, the friction force is directed to the left. The magnitude of the kinetic friction force is given by $$F_k = \mu_k F_N$$. For an object on a horizontal surface, the normal force $$F_N$$ equals the weight $$mg$$. Thus, the magnitude of the friction force is $$\mu_k mg$$ and is constant as long as the object is in motion and the coefficient of kinetic friction is constant.

The friction force is $$F_k = \mu_k F_N$$. First, we must find the normal force $$F_N$$. The vertical forces must sum to zero. Let the applied force be $$T$$. The upward forces are the normal force and the vertical component of the tension ($$T \sin\theta$$). The downward force is gravity ($$mg$$). So, $$F_N + T \sin\theta - mg = 0$$, which means $$F_N = mg - T \sin\theta$$. Using $$g \approx 9.8 , \text{m/s}^2$$, $$\sin 37° \approx 0.6$$: $$F_N = (20)(9.8) - (100)(0.6) = 196 - 60 = 136 , \text{N}$$. Now, calculate the friction force: $$F_k = (0.20)(136 , \text{N}) = 27.2 , \text{N}$$.

The maximum static friction force is given by $$F_{s,max} = \mu_s F_N$$. The area of the base of Block A is $$L^2$$ and for Block B is $$(2L)(L/2) = L^2$$. Although the areas are the same, the friction force is independent of the contact area. Since both blocks have the same mass, their weight $$mg$$ is the same, and the normal force $$F_N$$ from the horizontal surface is the same for both. They are made of the same material and are on the same surface, so $$\mu_s$$ is the same. Therefore, the maximum static friction force is the same for both blocks.