Calculating Stored Energy: A Tranquilizer Gun Spring

We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun?

The figure a shows an artistic impression of a tranquilizer gun, which shows the inside of it revealing the gun spring and a panel just below it, in the outside area, attached to the spring. This stage shows the gun before it is cocked, and the spring is uncompressed covering the entire inside area. The figure b shows the gun with the spring in the compressed mode. The spring has been compressed to a distance x, where x distance shows the vacant area inside the gun through which the spring has been compressed. The panel is also moving along the spring. And a bullet of mass m is shown at the front of the compressed spring. The spring here has elastic potential energy, represented by P E sub e l. The figure c is the third stage of the above two stages of the gun. The spring here is released from the compressed stage releasing the bullet in the outer forward direction with velocity V and the spring’s potential energy is converted into kinetic energy, represented here by K E.

Figure 16.7 (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance

x

, and the projectile is in place. (c) When released, the spring converts elastic potential energy

{PE}_{el}

into kinetic energy.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy for a
(a): The energy stored in the spring can be found directly from elastic potential energy equation, because $k$ and $x$ are given.
Solution for a
Entering the given values for $k$ and $x$ yields
$\begin{array}{l} {PE}_{el} & = & \frac{1}{2} {kx}^{2} = \frac{1}{2} (50.0 N/m) {(0.150 m)}^{2} = 0 . 563 N \cdot m \\ = & 0.563 J \end{array}$
16.5
Strategy for b
Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed.
4
Solution for b
1. Identify known quantities:
  ${KE}_{f} = {PE}_{el} or 1 / 2 {mv}^{2} = (1 / 2) {kx}^{2} = {PE}_{el} = 0 . 563 J$
  16.6
2. Solve for $v$ :
  $v = {[\frac{2 {PE}_{el}}{m}]}^{1 / 2} = {[\frac{2 (0.563 J)}{0.002 kg}]}^{1 / 2} = 23 . 7 {(J/kg)}^{1 / 2}$
  16.7
3. Convert units: $23.7 m / s$
Discussion
(a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.

0 of 5 steps open

How did it go?

Strategy for a

Solution for a

Strategy for b

Solution for b

Discussion