Question

If a dart gun with a 20N/m spring inside of it is compressed a distance of 0.3m, How high into the air can it shoot a 0.15kg dart?

Answer 1

Answer:

h = 61.16[cm]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that energy is conserved or equal in two points in space for an instant in time.

In this way we will have the points A & B, the point A for the moment before shooting and the moment B when the Dart is in the highest position.

In this way the energy is:

$E_{A}=E_{B}$

Now we must identify the energies in the moments A & B. in the instant A we have the spring compressed, in such a way that only elastic energy is stored.

$E_{A}=\frac{1}{2} *k*x^{2}$

where:

k = spring constant = 20 [N/m]

x = distance = 0.3 [m]

Now, at the moment when the dart is in the highest position (B), it means that it does not go up anymore, that is, its movement is zero, and therefore its kinetic energy is zero, in this way the energy at the highest point corresponds to potential energy.

$E_{B}=m*g*h$

where:

m = mass = 0.15[kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation [m]

Now replacing:

$\frac{1}{2} *20*(0.3)^{2}=0.15*9.81*h\\0.9=1.4715*h\\h=0.61[m]\\or\\h = 61.16[cm]$