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A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b=5.5 N•s2/m2 What is the terminal speed that the box will reach?

Respuesta :

yemmy

Answer:

1.03m/s

Explanation:

According to the question

Fdrag = bv²

now Fdrag = weight of the box

that is Fdrag = mg

where m = mass of the box, 0.6 Kg

and g = acceleration due to gravity

thus,

Fdrag = bv²   becomes

mg = bv²

making v the subject of formula

[tex]v = \sqrt{\frac{mg}{b} }[/tex]

[tex]v = \sqrt{\frac{0.6 * 9.8}{5.5} }[/tex]

v = 1.03m/s

Answer:

the terminal speed that the box will reach v = 1.0355 m/s

Explanation:

The drag force is the resultant force component that acts in the relative motion direction of the body.

Here we have Fdrag = bv²

But Fdrag = mg

Therefore F = 0.6 × 9.81 =5.886 N

v² = Fdrag / b = 5.886 N/(5.5 N s²/m²)

v = 1.0355 m/s

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