Give me a set of 5 difficult mechanics questions and their solutions.

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In this problem, we have a particle of mass m subject to a force that depends on its velocity squared. The force is given by F equals negative k times v squared, where k is a positive constant. The negative sign indicates that the force opposes the motion. The particle starts from rest at position x equals zero at time t equals zero. We need to find the velocity as a function of position, and the time it takes for the particle to reach a given position x. Let's solve for the velocity as a function of position. We start with Newton's second law, F equals m a. The acceleration can be written as dv over dt, so we have m times dv over dt equals negative k v squared. To relate velocity to position, we use the chain rule. Since v equals dx over dt, we can write a equals v times dv over dx. This gives us m times v times dv over dx equals negative k v squared. Dividing both sides by v, assuming v is not zero, we get m times dv over dx equals negative k v. This is a separable differential equation. We integrate both sides: the integral of dv over v equals negative k over m times the integral of dx. This gives us the natural logarithm of the absolute value of v equals negative k over m times x plus a constant C. Using the initial condition, at x equals zero, v equals v_0, we find that C equals the natural logarithm of v_0. Therefore, the natural logarithm of v over v_0 equals negative k over m times x. Solving for v, we get v of x equals v_0 times e to the power of negative k over m times x. This shows that the velocity decreases exponentially with distance. Now, let's find the time it takes for the particle to reach position x. We start with the definition of velocity: v equals dx over dt. We already found that v of x equals v_0 times e to the power of negative k over m times x. Substituting this expression, we get dx over dt equals v_0 times e to the power of negative k over m times x. Rearranging to isolate dt, we get dt equals dx divided by v_0 times e to the power of negative k over m times x, which equals e to the power of k over m times x divided by v_0 times dx. Now we integrate both sides. The left side integrates from 0 to t, and the right side from 0 to x. This gives us t equals 1 over v_0 times the integral from 0 to x of e to the power of k over m times x-prime dx-prime. Evaluating this integral, we get t equals 1 over v_0 times the quantity m over k times e to the power of k over m times x minus m over k. Simplifying, we get t of x equals m over k v_0 times the quantity e to the power of k over m times x minus 1. This shows that the time increases exponentially with distance, which makes sense because the particle slows down as it moves. In this problem, we have a uniform solid cylinder of mass M and radius R rolling without slipping down an inclined plane with angle theta. A string is wrapped around the cylinder's axle and passes over a pulley at the top of the incline. The string is connected to a hanging block of mass m. We need to find the acceleration of the block. This is a complex problem involving both rotational and translational motion. The cylinder experiences a gravitational force, tension from the string, and friction from the incline. The hanging mass experiences gravity and tension. Since the cylinder rolls without slipping, there's a relationship between its linear and angular accelerations. We'll need to apply Newton's laws and analyze the torques to solve this problem.