Question

A roller coaster has a mass of 650 kg. It sits at the top of a hill with height 78 m. If it drops from this hill, how fast is it going when it reaches the bottom? (Assume there is no air resistance or friction.)

Answer 1

Answer:

The roller coaster is going at 39.1 m/s when it reaches the bottom

Explanation:

We use conservation of mechanical energy to solve this problem. So, we need to estimate the potential energy of the cart at the top of the hill while it is sitting steady there. Such will give us the cart initial Total Energy (TE) since its kinetic energy at that moment is zero (the cart is not moving, just sitting at the top of the hill).

Its initial potential energy (Ui) is therefore: $Ui=m*g*h=650 * 9.8 * 78=496860 J$

Notice also that all units given are in the SI system, (including the acceleration of gravity g) and therefore the answer is in Joules.

As we said, this initial potential energy equals the Total Energy  (TE) of the roller coaster because its initial Kinetic Energy (Ki) is zero.

When we look at what happens the instant the cart reaches the bottom, at that point it has all kinetic energy and zero potential energy (h is zero at the bottom of the hill). Final Kinetic Energy is by definition: Kf =$\frac{1}{2} m*v^2=\frac{1}{2}650*v^2=325 v^2$

Therefore, the conservation of mechanical energy tells us that:

$ET_i=ET_F\\U_i+K_i=U_F+K_F\\496860 J +0J=0J+325v^2$

We can solve for the final velocity of the roller coaster in this equation by solving for it in the equation:

$496860 J +0J=0J+325v^2\\\\496860J=325v^2\\v^2=\frac{496860}{325} =1528.8\\v=\sqrt{1528.8} = 39.1 \frac{m}{s}$

The velocity should result with units of meters/second since all quantities were given in SI units.