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Rus_ich [418]
3 years ago
5

Bob is pushing a box across the floor at a constant speed of 1.4 m/s, applying a horizontal force whose magnitude is 65 N. Alice

is pushing an identical box across the floor at a constant speed of 2.8 m/s, applying a horizontal force. (a) What is the magnitude of the force that Alice is applying to the box? F = N (b) With the two boxes starting from rest, explain qualitatively what Alice and Bob did to get their boxes moving at different constant speeds. In order to keep the box moving twice as fast, Alice had to apply a constant force that was twice as large as the force that Bob applied. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.
Physics
1 answer:
bija089 [108]3 years ago
5 0

Answer:

a) 65 N

b) Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) Here, both boxes are identical. They have different velocities but the velocities are constant hence the force Alice is applying is 65 N. Constant velocity means there is no acceleration.

b) The acceleration that Alice provided must have been different than that of Bob. But after reaching their desired speed they stopped accelerating. This would make the net force zero.

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Answer:

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2) 3.31 seconds

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Explanation:

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1) Top speed = 28.7 m/s

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1\ m=\frac{1}{1609.344}\ miles

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28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h

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v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2

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v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s

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v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m

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4) When s = 120 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s

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