Asked by Aly
A man skis down a slope 120 m high. If 80 percent of his initial potential energy is lost to friction and air resistence, what is his speed at the bottom of the slope?
Answers
Answered by
drwls
20% of M g H is converted to (1/2) M V^2
Note that the mass M cancels out.
0.2 gH = 0.5 V^2
H = 120 m; g = 9.81 m/s^2
Now just Solve for V
Note that the mass M cancels out.
0.2 gH = 0.5 V^2
H = 120 m; g = 9.81 m/s^2
Now just Solve for V
Answered by
Anonymous
45
Answered by
marco
PE=KE
mgh=1/2mv^2
0.2(9.8m/s^2)(120m)=0.5(v^2)
(235.2m^2/s^2)/0.5=(0.5v^2)/0.5
v^2=470.4
v=21.689 m/s
mgh=1/2mv^2
0.2(9.8m/s^2)(120m)=0.5(v^2)
(235.2m^2/s^2)/0.5=(0.5v^2)/0.5
v^2=470.4
v=21.689 m/s
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.