A 119 kg man stands in a 141 kg rowboat at rest in still water. He faces the back of the boat and throws a 2 kg rock horizontally at a speed of 17 m/s. The boat recoils forward and comes to rest 2.32 m from its original position.

The acceleration of gravity is 9.8 m/s².

1. Calculate the initial recoil speed of the boat. Answer in units of m/s.

2. What is the loss in mechanical energy due to the frictional force exerted by the water? Answer in units of J.

3. What is the effective coefficient of friction between the boat and water?

To find the initial recoil speed of the boat, we can use the conservation of momentum principle. The total momentum before throwing the rock should be equal to the total momentum after throwing the rock.

The initial momentum is 0 (boat and person are at rest). After throwing the rock, the momentum of the rock is given by the mass of the rock (2 kg) times its velocity (17 m/s), which is 2*17 = 34 kg*m/s.

Let the initial recoil speed of the boat (with the person) be v. Then, the momentum of the boat after throwing the rock is (119 kg + 141 kg) * v. According to the conservation of momentum:
0 = 34 kg*m/s + (119 kg + 141 kg) * v
Rearranging the equation to solve for v, we get:
v = - 34 kg*m/s / (119 kg + 141 kg)
v ≈ -0.133 m/s
The negative sign indicates that the boat is moving in the opposite direction of the thrown rock. Therefore, the initial recoil speed of the boat is approximately 0.133 m/s.

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