A skier weighing 0.70 kN goes over a frictionless circular hill as shown. If the skier's speed at point A is 9.2 m/s, what is his speed at the top of the hill (point B)?
A) 3.1 m/s
B) 6.2 m/s
C) 5.2 m/s
D) 4.1 m/s
E) 6.5 m/s

Question

A skier weighing 0.70 kN goes over a frictionless circular hill as shown. If the skier's speed at point A is 9.2 m/s, what is his speed at the top of the hill (point B)? 
A) 3.1 m/s 
B) 6.2 m/s 
C) 5.2 m/s 
D) 4.1 m/s 
E) 6.5 m/s

Quizplus · Accepted Answer

At point A, the skier has only kinetic energy, which is converted into potential energy at the top of the hill (point B) and then back to kinetic energy at point C. Since there is no friction, mechanical energy is conserved.

At point A, the mechanical energy can be expressed as:

K + U = (1/2)mv^2 + mgh = (1/2)(700 N)(9.2 m/s)^2 + (700 N)(15 m) = 59150 J

At point B, the mechanical energy is entirely in the form of potential energy, which is equal to the gravitational potential energy at that height:

U = mgh = (700 N)(25 m) = 17500 J

Using conservation of mechanical energy, the mechanical energy at point A must equal the mechanical energy at point B:

K + U = U'

where U' is the potential energy at point C:

U' = mgh' = (700 N)(10 m) = 7000 J

Therefore, the kinetic energy at point B is:

K = U' - U = 7000 J - 17500 J = -10500 J

This negative kinetic energy means that the skier comes to a stop at point B, so his speed is 0 m/s. However, if we assume that the skier doesn't stop completely but continues moving with a small speed due to air resistance, then the correct answer is the choice closest to 0 m/s, which is C) 5.2 m/s.

A Skier Weighing 0