A 27g arrow is shot horizontally. The bowstring exerts an average force of 75N on the arrow over a distance of 78 cm. Determine, using the work energy theorem, the maximum speed reached by the arrow as it leaves the bow? Work = dfcostheta        Ef = 1/2mv^2

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The arrow that is being shot has a mass of 27 g. The bow string exerts a force of 75N over a distance of 78 cm. The work done by the bow string is stored in the arrow as kinetic energy. The kinetic energy that the arrow attains when it leaves...

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The arrow that is being shot has a mass of 27 g. The bow string exerts a force of 75N over a distance of 78 cm. The work done by the bow string is stored in the arrow as kinetic energy. The kinetic energy that the arrow attains when it leaves the bow is equal to (1/2)*m*v^2, where v is the maximum velocity.

The work done on the arrow by the bow string is given by force*displacement as the force acts in the same direction as the displacement of the arrow. Here cos theta = cos 0 = 1.

Work done = 75*78/100 = 58.5 J

The kinetic energy (1/2)*m*v^2 = 58.5

=> (1/2)*(27/1000)*v^2 = 58.5

=> v^2 = 58.5*2000/27

=> v^2 = 13000/3

=> v = 65.85 m/s

The arrow has the maximum velocity when it leaves the bow and it is equal to 65.85 m/s.

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