Question

A 89.35-kg skier with an initial speed of v_i=43.91 m/s coasts up a H=15.58-m-high rise as shown in the figure. The slope angle is theta=50.33. What is her final speed at the top? Assume the coefficient of friction between her skis and the snow is 0.069 (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) Use g = 10 m/s².

Answer 1

By Energy conservation:

KEi + PEi + Wfr = KEf + PEf

here, KEi = initial kinetic energy = 0.5*m*vi^2

PEi = initial potential energy = 0

KEf = final kinetic energy = 0.5*m*vf^2

PEf = final potentil energy = m*g*H

Wfr = work done by friction = Ff*d*cosA

where, Ff = friction force = *(Normal force) = *m*g*cos

d = distance travelled by skier = H/sin = 15.58/sin 50.33 deg = 20.24 m

given, = slope angle = 50.33 deg

H = height = 15.58 m

m = mass of skier = 89.35 kg

vi = initial speed = 43.91 m/s

= coefficient of friction = 0.069

g = 10 m/s^2

A = angle between friction force vector and displacement vector = 180 deg

vf = final speed at top = ??

So, 0.5*m*vi^2 + 0 + *m*g*cos*d*cosA = 0.5*m*vf^2 + m*g*H

Using known values:

0.5*89.35*vf^2 = 0.5*89.35*43.91^2 + 0.069*89.35*10*cos(50.33 deg)*20.24*cos(180 deg) - 89.35*10*15.58

vf = sqrt(43.91^2 + 2*0.069*10*cos(50.33 deg)*20.24*cos(180 deg) - 2*10*15.58)

vf = 39.98 m/sec. = 40.0 m/s