An archery bow is drawn a distance d = 0.29 m and loaded with an arrow of mass m = 0.094 kg. The bow acts as a spring with a spring constant of k = 112 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow. How fast will the arrow travel and it leave the bow?

The arrow will leave the bow with a velocity of 10 m/s.

Explanation:

Hi there!

The potential energy stored in the bow can be calculated using the following equation:

U = 1/2 · k · d²

Where

U = elastic potential energy.

k = spring constant.

d = stretched distance of the bow

Then:

U = 1/2 · 112 N/m · (0.29 m) ²

U = 4.7 J

When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:

KE = 1/2 · m · v² = 4.7J

Where:

KE = kinetic energy.

m = mass of the arrow.

v = velocity of the arrow:

Then:

4.7 J = 1/2 ·0.094 kg · v²

2 · 4.7 J / 0.094 kg = v²

9.4 kg · m²/s² / 0.094 kg = v²

v = 10 m/s

The arrow will leave the bow with a velocity of 10 m/s.