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emmainna [20.7K]
3 years ago
13

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

the shore and then stops. The boat has a mass of 46.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?
Physics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:16.096

Explanation:

Given

mass of dog\left ( m_d\right )=10kg

mass of boat\left ( m_b\right )=46kg

distance moved by dog relative to ground=x_d

distance moved by boat relative to ground=x_b

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0=m_d\times x_d+m_b\dot x_b

0=10\times x_d+46\dot x_b

x_d=-4.6x_b

i.e. boat will move opposite to the direction of dog

Now

|x_d|+|x_b|=7.8

substitutingx_dvalue

5.6|x_b|=7.8

|x_b|=1.392m

|x_d|=6.4032m

now the dog is  22.5-6.403=16.096m from shore

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Generally the force on this particle is mathematically represented as

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       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

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