Anew ride being built at an amusement park includes a vertical drop of 126.5 metered. 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 3.5 x 10^4 kg, how much potential energy was converted into thermal energy?

The amount of Potential Energy lost in the form of Thermal Energy is equal to 41.64 MJ.

Explanation:

mass of the cart and passengers = m = 3.5*10⁴ kg

Height = h = 126.5 m

g = 9.8 ms⁻²

v = 10 ms⁻¹

At the top position the Total energy of the ride is in the form of potential energy given as:

P.E = mgh

= 3.5*10⁴*9.8*126.5

= 43389500 J

At the bottom of the drop, all the Potential Energy will be converted into K.E as the ride develops speed given as:

K.E = 0.5*m*v²

= 0.5*3.5 x 10⁴*100

= 1750000 J

The K.E is much less than the P.E energy even though all the P.E of the ride has been converted. The loss of energy is due to the formation of thermal energy which can be calculated as:

Thermal Energy lost = Total energy in the form of P.E(top) - Total energy in the form of K.E(bottom)

= 43389500 -  1750000

= 41639500 J

= 41.64 M J approx             M J = mega joule = 10⁶ J

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