You have a 0.20 kg arrow and a 500.0 N/m bow.

Question

If you pull the bow back 40.0 cm, how fast will the arrow be moving if shot horizontally?

What will be the maximum height of the arrow if you shoot it straight up into the air with a stretch of 40.0 cm?

Answer 1

The bow is basically a spring (something elastic that exerts a force the more it's stretched/compressed)

So Force at every point is k*x where k is the 500N/m, the force is applied during the entire time the bow is in contact with the arrow so you have to integrate the force over the distance (this is calculus, if you're taking physics without calculus just look at the bold formula below) which will tell you work done by the force which is change in kinetic energy. So integrate: ∫ k*x*dx and get 0.5*k*x^2. That's the potential energy of the spring.

aka how much work the spring can do.

So.5*x^2 = 250(0.4 m)² = 40 joules. That will all be transferred to kinetic energy when the arrow leaves so set it equal to .5*m*v² to solve for v.

For part 2, use same velocity and work-kinetic-energy-theorem-formula: V²=V₀²-2gd and V² is zero because we're looking for where it shoots up and stops right before falling down so we're looking for where the final velocity is zero so V₀²/2(9.8) should be your answer.