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

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...