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 kinetic energy (KE) equal to mgh, where m is the mass of the skier, g is the acceleration due to gravity, and h is the height of the hill. At point B, the skier has potential energy (PE) equal to mgh. Since the hill is frictionless, there is no loss of mechanical energy, so the KE at point A is equal to the PE at point B, or KE = PE. Setting these equal to each other and solving for the speed at point B, we get:

1/2mv^2 = mgh
v^2 = 2gh
v = sqrt(2gh)

Plugging in the known values, we get:

v = sqrt(2*9.8*2.5) = 5.2 m/s

Therefore, the answer is C) 5.2 m/s.

A Skier Weighing 0