A sled drops 50 meters in height on a hill. The mass of the rider and sled is 70 kg and the sled is going 10 m/s at the bottom of the hill. What is the efficiency of energy conversion from potential to kinetic?

Efficiency = 10.2 %

Explanation:

Given the following data;

Mass = 70 kg

Height = 50 m

Velocity = 10 m/s

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the efficiency of energy conversion from potential to kinetic;

First of all, we would determine the potential energy;

P.E = mgh

P.E = 70 * 9.8 * 50

P.E = 34300 J

For the kinetic energy;

K.E = ½mv²

K.E = ½ * 70 * 10²

K.E = 35 * 100

K.E = 3500

Therefore, Input energy, I = 34300 J

Output energy, O = 3500 J

Next, we find the efficiency;

Efficiency = O/I * 100

Substituting into the formula, we have;

Efficiency = 3500/34300 * 100

Efficiency = 0.1020 * 100

Efficiency = 10.2 %

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