3. You (m-76kg) find yourself at Sunrise Ski Resort doing some snow skiing on yo
ID: 2035689 • Letter: 3
Question
3. You (m-76kg) find yourself at Sunrise Ski Resort doing some snow skiing on your retro wood skies (not modern waxed skies, m-2kg, and therefore there is some friction with a coefficient of friction of 0.06). On your way down a blue slope (40-degrees and 250m long) a child (m-27kg) skis out in front of you. You are half way down at this point and you don't have time turn, but you manage to scoop the child up (in an instant) and carry them the rest of the way down the slope. How fast are you and the child going when you reach the bottom?Explanation / Answer
from top of the hill to half the way down
work done by gravity Wg = (m + ms)*g*sintheta*L/2
work done by friction Wf = -u*(m+ms)*g*costheta*L/2
initial kE at the top KEi = 0
kinetic energy at half way KEf = (1/2)*(m + ms)*v1^2
from work energy relation
total work = chnage in KE
(m + ms)*g*sintheta*L/2 -u*(m+ms)*g*costheta*L/2 = (1/2)*(m + ms)*v1^2
(76+2)*9.8*sin40*250/2 - 0.06*(76+2)*9.8*cos40*250/2 = (1/2)*(76+2)*v1^2
v1 = 38.24 m/s
momentum before scooping Pi = (m + ms)*v1
momentum after scooping Pf = (m + ms + mc)*v2
from momentum conservation
Pf = Pi
(m + ms + mc)*v2 = (m + ms)*v1
(76 + 2 + 27)*v2 = (76 + 2)*38.24
v2 = 28.4 m/s
from half way to down
work done by gravity Wg = (m + ms + mc)*g*sintheta*L/2
work done by friction Wf = -u*(m + ms + mc)*g*costheta*L/2
KE at half way = (1/2)*(m + ms + mc)*v2^2
KE at the bottom KEf = (1/2)*(m + ms + mc)*v3^3
work = change in KE
(m + ms + mc)*g*sintheta*L/2 -u*(m + ms + mc)*g*costheta*L/2 = (1/2)*(m + ms + mc)*(v3^2 - v2^2)
(76 + 2 + 27)*9.8*sin40*250/2 - 0.06*(76 + 2 + 27)*9.8*cos40*250/2 = (1/2)*(76 + 2 + 27)*(v3^2 - 28.4^2)
v3 = 47.63 m/s <<<----------ANSWER
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.