In a downhill ski race surprisingly little advantage is gained by getting a runn

Question

In a downhill ski race surprisingly little advantage is gained by getting a running start. This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills. To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a 25° slope neglecting friction for the following two cases. (Note that this time difference can be very significant in competitive events so it is still worthwhile to get a running start.)

(a) starting from rest final speed

______m/s

time taken

_______s

(b) starting with an initial speed of 3.00 m/s final speed

________m/s

time taken

________s

Explanation / Answer

, find the final speed and the time taken for a skier who skies 70.0 m along a 25° slope neglecting friction for the following two cases.

Length of slope, d = 70 m
Angle of slope, A = 25

Part a:
From Work Energy Theoram,
Work Done = KInetic Energy
Force * Displacement * Cos0 = 0.5 * m * v^2
mgSin25*d = m*v^2

solving for Velocity, v

v = sqrt(gSin25*d)
v = sqrt(9.81*sin25*70)
v = 17.036 m/s

also from newton Second law,
a = F/m
a = mgSin25/m
a = gSin25
a = 9.81*sin25
a = 4.146 m/s^2

From First Eqn of motion,
time period , t = v/a
t = 17.036/4.146
t = 4.109 s

Part B:
From Work Energy Energy Theoram
work done + initial KE = Final KE
F*d*Cos0 +0.5 * m * vi^2 = 0.5*m*v^2

solving for final velocity, v

v = sqrt(gsin25*d + vi^2)
v = sqrt(9.81*sin25*70 + 3^2)
v = 17.298 m/s

also from newton Second law,
a = F/m
a = mgSin25/m
a = gSin25
a = 9.81*sin25
a = 4.146 m/s^2

From First Eqn of motion,
time period , t = (v-u)/a
t = (17.298 - 3)/4.146
t = 3.449 s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.