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hoa [83]
3 years ago
6

A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the

friction force that act on the block.​
Physics
2 answers:
mars1129 [50]3 years ago
6 0

Answer:

FAE= 0.014 N

Explanation:

The KE of block is decreased because of the slowing action of the friction force .

Change in KE of block = work done on block by friction ƒ

⠀ ➪ ½mu²ƒ - ½mu²i = Fƒs cos θ

Because the friction force on the block is opposite in direction to the displacement , cos θ = -1

➢ Using Uƒ = 0 , Vƒ = 0.20 m/s , and s = 0.70 m

✒ We find ,

➪½mu²ƒ - ½mu²i = Fƒs cos θ

➪0-½ (0.50 kg) (0.20 m/s)² = (Fƒ) (0.70 m) (-1)

➪ Fƒ = 0.014 N

Hope this helped, can i pls have brainliest

Alexeev081 [22]3 years ago
5 0
0.014 is the answer. Hope that helps!
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