Gayle runs at a speed of 3.45 meter per second and dives on a sled, initially at rest on the top...

Question:

Gayle runs at a speed of {eq}3.45 {/eq} meter per second and dives on a sled, initially at rest on the top of a friction less snow-covered hill. After she has descended a vertical distance of {eq}5.00 m {/eq}, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is {eq}15.0 m {/eq}? (Gayle's mass is {eq}52.0 kg {/eq}, the sled has a mass of {eq}5.40 kg {/eq} and her brother has a mass of {eq}30.0 kg {/eq})

Potential Energy:

When doing energy conservation calculations, potential energies and kinetic energies are usually added to get the total energy. This total energy does not change whatever happens. Increases in kinetic energy are accompanied by decreases in potential energy, and vice-versa. When dealing with classical physics problems (kinematics), most of the potential energies are due to gravitation. In these cases, the potential energy can be expressed as:

{eq}\displaystyle U = mgh {/eq}

where:

• m is the mass of an object
• h is its height from the ground
• {eq}\displaystyle g = 9.81\ m/s^2 {/eq} is the acceleration due to gravity

Given:

• {eq}\displaystyle v_1 = 3.45\ m/s {/eq} is the speed of Gayle
• {eq}\displaystyle h_2 = 10\ m {/eq} is the height at which her brother joins her
• {eq}\displaystyle h_1 = 15\ m {/eq} is the height at which Gayle jumps on the sled
• {eq}\displaystyle m_1 = 52\ kg {/eq} is the mass of Gayle
• {eq}\displaystyle m_2 = 5.4\ kg {/eq} is the mass of the sled
• {eq}\displaystyle m_3 = 30\ kg {/eq} is the mass of her brother

In order to solve this problem, we can use the conservation of energy. At the initial stage of the situation, only Gayle is moving. Therefore, only she has a kinetic energy. The sled has a gravitational potential energy since it is initially at rest at 15 m above, and her brother is initially at rest at 10 m above the bottom point.

As for the last stage of the situation when they are all moving together, we have one kinetic energy term with the speed we are looking for, and the masses of all three together. So we set up our equation as:

{eq}\displaystyle \frac{1}{2} m_1 v_1^2 + m_2 g h_1 + m_3 g h_2 = \frac{1}{2} (m_1 + m_2 + m_3)v^2 {/eq}

let us substitute all the known variables:

{eq}\displaystyle \frac{1}{2} (52\ kg)(3.45\ m/s)^2 + (5.4\ kg)(9.81\ m/s^2)(15\ m) + (30\ kg)(9.81\ m/s^2)(10\ m) = \frac{1}{2} (52\ kg + 5.4\ kg + 30\ kg)v^2 {/eq}

we simplify:

{eq}\displaystyle 309.465\ J + 794.61\ J + 304.3\ J = (43.7\ kg)v^2 {/eq}

{eq}\displaystyle 1408.375\ J = (43.7\ kg)v^2 {/eq}

we isolate the speed:

{eq}\displaystyle v = \sqrt{\frac{1408.375\ J}{43.7\ kg}} {/eq}

we will thus obtain:

{eq}\displaystyle \boxed{v = 5.68\ m/s} {/eq} 