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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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B. 4 m/s

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Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

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I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

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3 years ago
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Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

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(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

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V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

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