A 28 kg canoe has a 24 kg passenger. The passenger is 2.0 m from the left end of the canoe and the center of mass of the canoe by itself is 2.9 m from the left end.

a. Calculate the position of the center of mass of the canoe/passenger system. b. if we neglect any horizontal forces exerted by the water on the canoe and we assume there is no wind that day, what is the next horizontal force on the canoe/passenger system?
c. Under the conidtions of (b), what is the acceleration of the center of mass of the canoe/passenger system?
d. Under the conditions of (b), the passenger moves to a point 1.0 to the right of the center of mass of the canoe. By how much does the position of the center of mass of the canoe/passenger system change during this process with respect to the the river bank?

a. To calculate the position of the center of mass of the canoe/passenger system, we need to take into account the mass and position of both the canoe and the passenger.

The center of mass is the weighted average of the positions of the individual masses involved, weighted by their respective masses. In this case, we have:

Canoe mass (m1) = 28 kg
Canoe position (x1) = 2.9 m from the left end

Passenger mass (m2) = 24 kg
Passenger position (x2) = 2.0 m from the left end

To calculate the position of the center of mass (x_cm), we can use the following formula:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Substituting the given values, we have:

x_cm = (28 kg * 2.9 m + 24 kg * 2.0 m) / (28 kg + 24 kg)
= (81.2 kg·m + 48 kg·m) / 52 kg
= 129.2 kg·m / 52 kg
= 2.48 m

Therefore, the position of the center of mass of the canoe/passenger system is 2.48 m from the left end.

b. If we neglect any horizontal forces exerted by the water or wind on the canoe/passenger system, the only horizontal force acting on it would be the force of the next horizontal force on the system. However, since there are no other forces mentioned in the given information, it is not possible to determine the magnitude or direction of this force without additional information.

c. Under the conditions mentioned in (b), the acceleration of the center of mass of the canoe/passenger system would be zero, as there are no horizontal forces present to cause acceleration.

d. If the passenger moves to a point 1.0 m to the right of the center of mass of the canoe, we need to calculate how much the position of the center of mass of the canoe/passenger system changes with respect to the river bank.

The change in position of the center of mass (delta_x_cm) would be equal to the change in position of the passenger (delta_x2), as the position of the canoe remains unchanged.

delta_x_cm = delta_x2

Since the passenger moves 1.0 m to the right of the center of mass (2.48 m), we have:

delta_x_cm = delta_x2 = 1.0 m

Therefore, the position of the center of mass of the canoe/passenger system changes by 1.0 m with respect to the river bank.