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A 0.53 kg arrow leaves a bowstring at a velocity of 63 m/s. If the arrow was initially at rest and then the string applied a force on it for 7 x 10-3 seconds, what was the approximate average force in Newtons that acted on the arrow during this time?

Respuesta :

lucic

Answer:

4770 N

Explanation:

Momentum is the product of mass and velocity and force is the change in momentum divided by change in time.

Given from the question;

Mass of arrow= 0.53 kg

Velocity of arrow = 63 m/s

Initial velocity of arrow = 0 m/s

Change in time = 0.007 s

Finding momentum after the arrow is released as;

p=m*v

p= 0.53 * 63

p= 33.39 kg*m/s

Force is the change in momentum divided by change in time;

F= 33.39 / 0.007

F= 4770 N

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