### Video Transcript

An archer pulls the string of a bow back by 0.50 metres, creating a tension of 150 newtons in the string. The mass of the arrow is 0.050 kilograms and the bowstring’s mass is negligible. What is the speed of the arrow after the bowstring is released and has returned to its equilibrium length? Assume that the force of the stretched bowstring on the arrow behaves like the force exerted by an extended spring and assume air resistance is negligible.

Let’s highlight some of the important information we’ve been given. We’re told the bowstring is pulled back by a distance of 0.50 metres and that this creates a tension of 150 newtons in the string. The mass of the arrow is given as 0.050 kilograms. We want to know the speed of the arrow after it’s been released. We’ll call that 𝑣.

Let’s begin our solution by drawing a diagram of the situation. Our diagram for this action has two parts: the first is a snapshot taken when the bow is stretched fully back and the second snapshot is taken after the arrow has been released and the bowstring has returned to its equilibrium position. In the first instance, snapshot one, the bowstring is pulled back a distance of 𝑥 metres; that’s, 0.50 metres. And we’re told that a tension 𝑇 of 150 newtons is established in the bow. The mass of the arrow, which we’ve called 𝑚 sub 𝑎, is given as 0.050 kilograms. And we want to know the speed of the arrow after it is released.

Let’s recall the relationship for the force associated with a compressed or stretched spring. Spring force 𝐹 sub 𝑠 is equal to the negative of the spring constant 𝑘 times the spring’s displacement from equilibrium 𝑥. The negative sign simply indicates that the force is in the opposite direction of the displacement. Looking at snapshot one, the displacement 𝑥 is to the left, but the spring force that the bowstring exerts on the arrow is to the right in the opposite direction.

If we ignore this direction or difference, we can write the spring force equation for our scenario. The tension 𝑇, which is created by the spring force, is equal to the spring constant of the bowstring system multiplied by the displacement 𝑥. We’re given the tension 𝑇 and displacement 𝑥, we can use these to solve for 𝑘. Let’s divide through both sides by 𝑥 to isolate 𝑘 on one side.

With 𝑥 cancelled from the right-hand side, we find that the spring constant 𝑘 is equal to the tension 𝑇 divided by 𝑥. Plugging in for 𝑇 and 𝑥, 150 newtons divided by 0.50 metres is equal to 300 newtons per metre. Now that we know the affected spring constant of the bow and string, let’s recall two important relationships that will help us solve for 𝑣.

First, let’s remember that potential energy stored up in a spring. That potential energy is one-half the spring constant multiplied by the displacement from equilibrium squared. The second fact we can recall is that in any closed system energy is conserved. The law of conservation of energy tells us that for any isolated system the initial energy at that system is equal to the final energy at that system at some later time.

For our bowstring problem, conservation of energy says that the energy of the system in snapshot one is the same as the energy of the system in snapshot two. In snapshot one, the system has only one type of energy, spring potential energy, expressed as one-half 𝑘𝑥 squared. In snapshot two, all the spring potential energy is released and converted into the kinetic energy of the arrow. We can recall that an object’s kinetic energy is equal to one-half its mass times its speed squared. Therefore, one-half 𝑘𝑥 squared in snapshot one equals one-half 𝑚 sub 𝑎 𝑣 squared.

In snapshot two, we want to solve for 𝑣, the speed of the arrow. First, we notice that the one-half terms on each side cancel. If we divide both sides by 𝑚 sub 𝑎 causing that term to cancel out on the right side and then take the square root of both sides causing the squared term to cancel with the square root on the right side, we find that 𝑣, the speed of the released arrow, is equal to the square root of the spring constant 𝑘 times 𝑥 squared divided by the mass of the arrow.

We’ve either been given or solved for each of these three variables. When we plug them in and compute this square root on our calculator, we find a value for 𝑣 to two significant figures of 39 metres per second. That’s how fast the arrow would leave the bow after the bowstring has returned to its equilibrium position.