prestomiller1234
Answered

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A loaded 500 kg sled is traveling on smooth horizontal snow at 5 m/s when it suddenly comes to a rough region. The region is 10 m long and reduces the sled's speed to 3 m/s what average friction force did the rough region exert on the sled?

Sagot :

masbrosur

Answer:

400 N

Explanation:

Change of Kinetic Energy to Friction Wok

∆KE = W

½ x m x (v(5)² - v(3)²) = f x d

½ x 500 x (5² - 3²) = f x 10

250 x (25 - 9) = f x 10

25 x 16 = f

f = 400 N

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