Question

A new ride being built at an amusement park includes a vertical drop of 71.6 meters. Starting from rest the ride vertically drops that distance before the track curves forward. If the velocity at the bottom of the drop is 10.0 m/s and the mass of the cart and passengers is 3,5x10^4kg, how much potential energy was converted into thermal energy? 2.3x10^7, 2.6x10^7, 4.3x107, 6.7x10^7

Answer 1

Answer:

We can conclude 2.3x10^7 J was converted to thermal energy

Explanation:

Energy Conservation

According to the law of conservation of energy, the total energy of an isolated system must be constant. If some kind of energy is 'lost', we know it was transformed into another type.

Let's check the conditions of the problem. The ride vertically drops a distance of 71.6 m starting from rest, and at the bottom of the drop, its speed is 10 m/s. Knowing the mass of the cart plus passengers is 3.5X10^4 kg, we compute the total energy at the top of the drop.

$E=U+K$

Where E is the total energy (which must be conserved) at the top of the drop, U is the gravitational potential energy and K is the kinetic energy. We use the equations for each:

$U=m.g.h$

$\displaystyle K=\frac{mv^2}{2}$

$\displaystyle E=m.g.h+\frac{mv^2}{2}$

At the top, the speed is 0, thus

$\displaystyle E=m.g.h=(35000)(9.8)(71.6)=2.5X10^7\ J$

Now we compute the 'total' energy at the bottom (quoted because we know there is some mechanical energy loss in the drop)

$\displaystyle E'=m.g.h'+\frac{mv'^2}{2}$

This time h'=0 and v=10 m/s, thus

$\displaystyle E'=\frac{(35000)10^2}{2}=1.8x10^6\ J$

The mechanical energy at the top and the bottom are not the same, thus we can know part of it was converted to heat or thermal energy. We compute the difference

$2.5x10^7\ J-1.8x10^6\ J=2.3x10^7\ J$

We can conclude 2.3x10^7 J was converted to thermal energy