An archer pulls back 0.75 m on a bowwhich has a stiffness of 200 N/m.

1.	An archer pulls back 0.75 m on a bowwhich has a stiffness of 200 N/m. Thearrow weighs 50 g. What is the velocityof the arrow immediately after release?
Answer» This can be solved using an energy method. We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow. The bow can be treated as a type of spring. The potential energy of a spring is: (1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed. Therefore, the potential energy PE of the bow is: PE = (1/2)(200)(0.75)2 = 56.25 J The kinetic energy of a particle is: (1/2)mv2, where m is the mass and v is the velocity. The arrow can be treated as a particle since it is not rotating upon release. Therefore, the kinetic energy KE of the arrow is: KE = (1/2)(0.05)v2 If we assume energy is conserved, then PE = KE Solving for the velocity of the arrow v we get v = 47.4 m/s

An archer pulls back 0.75 m on a bowwhich has a stiffness of 200 N/m. Thearrow weighs 50 g. What is the velocityof the arrow immediately after release?

Answer»

This can be solved using an energy method.

We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow.

The bow can be treated as a type of spring. The potential energy of a spring is:

(1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed.

Therefore, the potential energy PE of the bow is:

PE = (1/2)(200)(0.75)2 = 56.25 J

The kinetic energy of a particle is:

(1/2)mv2, where m is the mass and v is the velocity.

The arrow can be treated as a particle since it is not rotating upon release.

Therefore, the kinetic energy KE of the arrow is:

KE = (1/2)(0.05)v2

If we assume energy is conserved, then

PE = KE

Solving for the velocity of the arrow v we get

v = 47.4 m/s

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