Please show work. Thanks Question 4 (3 points) A 50-kg sledder is sliding North

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Please show work. Thanks

Question 4 (3 points) A 50-kg sledder is sliding North on a large frozen lake at 2 m/s, when a 30-s gust of wind from East to West blows her off course. 90 s after the wind stops (two minutes after the start of the journey), the sledder is 339.4 m North-West of her initial position (when the wind started) If the sledder gets no other pushes other than the wind and just slides freely otherwise during the two minutes, what was the magnitude of the average force exerted on her by the Wind (2 S.f.) Your Answer: Answer units Hide hint for uestion 4 Assume constant acceleration (MODEL) from East to West while the wind is blowing in order to work out the acceleration. You can use a force analysis or a momentum analysis to calculate the average force exerted by the wind.

Explanation / Answer

Initial velocity, u = 2 j [ j is towards noth, i is towards east]
The gust of wind is in -i direction , so the j component of velocy would remain unchanged
Velocity after 30 sec. gust of wind = ai + 2j
Final position = 339.4[-i + j]/sqroot(2)

Now, since after the wind there is no extra force, the y component of velocity would be responsible for the y displacement
2*120 = 339.4/sqroot(2) -- ( which is true)

to find the position before 90 seconds, -339.4/sqroot(2) = a*90
a = -2.667 m/s

so, v after gust = -2.667i + 2j
v before gust = 2j
change in velocity = -2.667 i
change in momentum = -2.667*50 i = -133.33 i = impulse = Fav*30
Fav = -4.444 i N

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