## Conservation of Energy

A sling shot, which has a band stiffness of 125 N/m, is pulled back 0.15m. If the pellet/rock weighs 100 grams, what is the velocity of the pellet/rock immediately after release?

##
__
__**Hint**

**Hint**

Kinetic Energy:

$$$KE=\frac{1}{2}mv^2$$$

where
$$m$$
is the mass, and
$$v$$
is the velocity.

##
__
__**Hint 2**

**Hint 2**

Potential Energy:

$$$PE=\frac{1}{2}kx^2$$$

where
$$k$$
is the spring constant/stiffness, and
$$x$$
is the spring’s change in length from un-deformed to deformed.

This problem can be solved using the conversion of Potential Energy into kinetic. Because the sling shot acts like a spring, the potential energy is:

$$$PE=\frac{1}{2}kx^2$$$

where
$$k$$
is the spring constant/stiffness, and
$$x$$
is the spring’s change in length from un-deformed to deformed.

$$$PE=\frac{1}{2}\cdot 125\frac{N}{m}\cdot (0.15m)^{2}=1.4\:J$$$

Once released, the pellet/rock will be converted to Kinetic Energy:

$$$KE=\frac{1}{2}mv^2$$$

where
$$m$$
is the mass, and
$$v$$
is the velocity.

$$$KE=\frac{1}{2}(0.1kg)v^2=(0.05)v^2$$$

Assuming energy is conserved, then
$$PE=KE$$
:

$$$1.4\:\frac{kg\cdot m^2}{s^2}=(0.05\:kg)v^2$$$

$$$v^2=28\:\frac{m^2}{s^2}\Rightarrow v=\sqrt{28m^2/s^2}=5.3\:m/s$$$

5.3 m/s