Question #5153

A .72 kg arrow is pulled back in a bow and pointed straight up into the air. The arrow has 162 J of elastic potential energy. How fast will the arrow be moving when it is released from the bow?

Expert's answer

According to the** energy conservation law** the potential elastic energy is transformed into the kinetic energy:

Ep = (m*V^2)/2. So,

V = sqrt(2*Ep/m) = sqrt(2*162/0.72) = 21.2 (m/sec).

Ep = (m*V^2)/2. So,

V = sqrt(2*Ep/m) = sqrt(2*162/0.72) = 21.2 (m/sec).

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