A bow has a spring constant of 388 N/m. It is pulled back 0.860 m, and used to fire a 0.105 kg arrow. When it is 45.4 m above the ground, how much KE does it have?

Answer 1

Let

k = 388 N/m be the spring constant,

d = 0.86 m be the displacement of the bow,

m = 0.105 kg be the mass of the arrow,

h = 45.4 m be the height reached by the arrow,

g = 9.8 m/s² be gravitational acceleration.

The energy of the bow, kd²/2,

gets converted to the arrow's potential energy, mgh,

plus the kinetic energy KE to be calculated.

Therefore

KE = kd²/2 - mgh

Substituting actual numbers

KE = 388 x 0.86²/2 - 0.105 x 9.8 x 45.4 = 96.7658 J