Question

When an archer pulls an arrow back in his bow, he is storing potential energy in the stretched bow. (a) Compute the potential energy stored in the bow, if the arrow of mass 5.05 10-2 kg leaves the bow with a speed of 32.0 m/s. Assume that mechanical energy is conserved. (b) What average force must the archer exert in stretching the bow if he pulls the string back a distance of 30.0 cm?

Answer 1

(a) If mechanical energy is conserved, then the kinetic energy of the arrow after it is shot equals the

potential energy of the arrow before it is shot.

kinetic energy is 1/2mv^2

potential energy is U so

1/2( 5.05*10^-2 ) (32)^2 = U

U = 25.856 J of potential energy

(b) I think you can use the equation for the potential energy of a spring for this part

U = 1/2 k x^2

25.856 = 1/2 k .3^2

k = 574.578

then you use the spring constant to fink the force necessary to stretch the bow

F = kx

F = 574.578 x .3

F = 172.37 N