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?

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, wherekis the stiffness andxis the amount the spring is stretched, or compressed.
Therefore, the potential energyPEof the bow is:
PE= (1/2)(200)(0.75)²= 56.25 J
The kinetic energy of a particle is:
(1/2)mv², wheremis the mass andvis the velocity.
The arrow can be treated as a particle since it is not rotating upon release.
Therefore, the kinetic energyKEof the arrow is:
KE= (1/2)(0.05)v²
If we assume energy is conserved, then
PE=KE
Solving for the velocity of the arrowvwe get
v= 47.4 m/s

sher mohammad

b.tech iit delhi