Question

An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after release?​

Answer 1

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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bolne se phle socha krr

Answer 2

Answer:

hey mate...

here is your answer...

Let us consider the potential energy of the bow to the kinetic energy of the arrow.

The bow can be assumed as a type of spring.

Therefore, the potential energy of a spring is:

begin mathsize 14px style 1 half kx squared end style, where 'k' is the stiffness and 'x' is the amount the spring is stretched, or compressed.

Hence, the potential energy P.E. of the bow is,

P.E. =begin mathsize 14px style equals 1 half cross times 200 cross times left parenthesis 0.75 right parenthesis squared equals 56.25 space straight J end style

Kinetic energy of a particle is,

begin mathsize 14px style 1 half mv squared end style, where 'm' is the mass, 'v' is the velocity.

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

begin mathsize 14px style therefore end styleThe kinetic energy K.E. of the arrow is,

begin mathsize 14px style straight K. straight E. equals 1 half cross times 0.05 cross times straight v squared space space space space space space space space equals space 0.025 space straight v squared end style

If we assume that the energy is conserved, then

P.E. = K.E.

So, the velocity of the arrow 'v', we get,

56.25 = 0.025 v2

v2=2250

v = 47.4 m/s

hope it helps...