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sdas [7]
3 years ago
14

A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant

Physics
1 answer:
GREYUIT [131]3 years ago
5 0

Your diagram should include four forces:

• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)

• the normal force, pointing up (mag. <em>n</em>)

• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")

• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

<em>n</em> + (-<em>w</em>) = 0

and

<em>p</em> + (-<em>f</em> ) = 0

So then the forces have magnitudes

<em>w</em> = 43.2 N

<em>n</em> = <em>w</em> = 43.2 N

<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

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Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

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Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

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