Question

A dog weighing 5.0 kg is standing on a flatbed boat so that he is 20.0 m from the shore. He walks 8.0 m on the boat toward the shore and then stops. The boat has a mass of 20 kg, and one can assume there is no friction between it and the water. How far is he from the shore at the end of his walk? Why?

Answer 1

Question

The dog will push the boat from the shore:

$$mgS_{dog} = MgS_{boat} \Rightarrow S_{boat} = \frac{mgS_{dog}}{Mg} = \frac{mS_{dog}}{M} = \frac{5 \cdot 8}{20} = 2 \text{ m}.$$

So, the boat will go away from shore by 2 meters. As result the distance from dog to shore will be:

$$d = 20 - 8 + 2 = 14 \text{ m}.$$

He will push the boat because of friction between dog’s paws and boat.

**Answer:** the dog will be 14 meters away from the shore.