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Leya [2.2K]
3 years ago
12

When Bella pushes a box with a mass of 5.25 kilograms with a force of 15.75 newtons, it accelerates at a rate of 2.5 meters/seco

nd2. What is the force due to friction?
I have no clue whats going on. Can anybody help me please!?!
Physics
2 answers:
Mariana [72]3 years ago
5 0

Sure !

Start with Newton's second law of motion:

                     Net Force = (mass) x (acceleration) .

This formula is so useful, and so easy, that you really
should memorize it.

Now, watch:

The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of  2.5 m/s² .
What's the net force on the box ?

                    Net Force = (mass) x (acceleration)

                                     = (5.25 kilograms) x (2.5 m/s²)

                     Net force =       13.125 newtons .

But hold up, hee haw, whoa !  Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?

==>  Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.

How much ?

            (Bella's 15.75 newtons) minus (13.125 that the box feels)

           =      2.625 newtons backwards, applied by friction.


dem82 [27]3 years ago
4 0

Answer:

-2.60

Explanation:

Plato

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G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

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Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

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y=y_{0}+V t

y=\frac{1}{2}-40 t

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Therefore the distance between them is given by,

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The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

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Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

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