Question

“What would be the kinetic energy of a skier at the foot or a 30 degree incline 39.2 meters long, his mass being 70 kilograms, had his final velocity been decreased 10% by friction?”

Answer 1

Height of the incline. h = xsinθ = (39.2 m) x sin 30^o = 19.6 m

Apply energy conservation law:

Initial potential energy = final kinetic energy + work done by friction

Let "V" be the actual final velocity.

Hence, Final velocity = 0.9V

mgh = 1/2.(m).(V²) + 1/2.(m).(0.1V)²

gh = 1/2 (V² + 0.1V²)

(19.6) x (9.8) = (0.5) x (1.1V²)

V² = (2 x 9.8 x 19.6) / (1.1)

V² = 349.2

V = 18.69 m/s

Hence, Final speed of the skier = (0.9) x (18.69 m/s) = 16.82 m/s

Therefore, Kinetic energy of the skier at the end is given by:

K.E = 1/2mv²

K.E = (0.5) x (70 kg) x (16.82 m/s)²

K.E = 9900 J --------------- (**Answer**)

(in case of anything wrong/have any doubts, please reach out to me via comments. I will help you)