physics help?

Question

An archer puts a 0.2 kg arrow to the bowstring. An average force of 189.8 N is exerted to draw the string back 1.44 m. g=9.8 m/s/s and assume no friction loss. With what speed does the arrow leave the bow (in m/s) and how high does it go if it is shot straight up (m)? No time is given.

Answer 1

The average of hthe force req'd to pull the string is 189.8 N. The string is pulled back 1.44 m.
So the energy stored in the string = 189.8 N * 1.44 m

That energy is converted completely to kinetic as it leaves the string. It is converted completely to potential at the top of its vertical flight.

So KE = 1/2 mv^2 = energy stored in the string
and U = m*g*h = energy stored in the string.