Question

An arrow with a mass of 250 g has a speed of 27 m/s before it passes through a 1.5 cm piece of plywood, after which it emerges at a speed of 15 m/s. What force did the plywood put on the arrow?

Answer #1

given M_{A} = Mass of arrow = 250 g = 0.250 Kg ,
V_{i} = Velocity of arrow before impact = 27 m/s

T = d = thickness of plywood piece = 1.5 cm = 0.015 m

V_{f} = velocity of arrow after impact = 15 m/s

We need work - energy principle to solve this problem

Work done = Change in kinetic energy

=> Kinetic Energy before impact = Kinetic energy after impact + work done by frictional force

=> 91.125 = 28.125 + F_{f} x 0.015

=> 91.125 - 28.125 = F_{f} x 0.015

=> 63 = F_{f} x 0.015

=> F_{f} = 63 / 0.015 = 4200 N

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