A)A 100kg person is riding a 10kg bicycle up a 25degree hill. The hill is long and the coefficient of static friction is 0.9. How much force does the person have to apply in order for her to ride 10m up the hill at constant velocity? B) If the person now starts from rest from the topf of the hill and rolls down a distance of 7m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?

but, but, it would have been fun :(

To find the force the person needs to apply to ride up the hill at a constant velocity, we need to consider the forces acting on the system.

A) The forces acting on the person and the bicycle going up the hill are the force of gravity (mg), the normal force (Fn), and the force of friction (Ff).

The force of gravity can be calculated using the mass of the person (m) and the acceleration due to gravity (g). In this case, the mass is 100 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So, the force of gravity (Fg) is Fg = m * g = 100 kg * 9.8 m/s^2 = 980 N.

The normal force (Fn) is the force exerted by the hill perpendicular to the surface. It is equal in magnitude but opposite in direction to the force of gravity, so Fn = -980 N.

The force of friction (Ff) can be calculated using the coefficient of static friction (μs) and the normal force (Fn). The formula for static friction is Ff = μs * Fn. In this case, the coefficient of static friction is 0.9, and the normal force is -980 N. So, Ff = 0.9 * (-980 N) = -882 N.

To ride up the hill at a constant velocity, the person needs to apply a force equal in magnitude but opposite in direction to the force of friction. So, the force the person needs to apply is 882 N.

B) To find the work done by friction to bring the bicycle to a full stop, we need to consider the forces acting on the system.

The forces acting on the bicycle going down the hill are the force of gravity (mg), the normal force (Fn), and the force of kinetic friction (Ff).

The force of gravity and the normal force are the same as in part A because they depend only on the mass of the person and the acceleration due to gravity. So, Fg = 980 N and Fn = -980 N.

The force of kinetic friction (Ff) can be calculated using the coefficient of kinetic friction (μk) and the normal force (Fn). The formula for kinetic friction is Ff = μk * Fn. In this case, the coefficient of kinetic friction is 0.65, and the normal force is -980 N. So, Ff = 0.65 * (-980 N) = -637 N.

When the person squeezes hard on the brakes and locks the wheels, the force of kinetic friction opposes the motion and brings the bicycle to a stop. The work done by friction is given by the formula W = F * d * cos(θ), where F is the force, d is the distance, and θ is the angle between the force and the direction of motion.

In this case, the force of kinetic friction is -637 N and the distance is 7 m. The angle between the force and the direction of motion is 180 degrees because they are opposite directions. So, W = -637 N * 7 m * cos(180 degrees) = -637 N * 7 m * (-1) = 4,459 J (Joules).

Therefore, the work done by friction to bring the bicycle to a full stop is 4,459 Joules.

Why are you posting a connexus question? You know you can be kicked out for cheating.