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2 months ago

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A 10 cm wide box is held between two springs in a 1 m gap on a frictionless surface. The left spring has a natural length of 80

cm and spring constant of 200 N/m. The right spring has a natural length of 90 cm and spring constant of 350 N/m. How far is the center of the box from the left edge in m

1 answer:

Maru [3.3K]2 months ago

6 0

Answer:

The distance measures $x =0.291 \ m$

Explanation:

According to the problem statement,

The box's width is $b = 10 \ cm = \frac{10}{100} = 0.10 \ m$

There is a gap of length $l = 1\ m$

The first spring's natural length is $y = 80 \ cm = \frac{80 }{100} = 0.8 \ m$

The spring constant for the first spring is $k_1 = 200 N/m$

The second spring has a natural length of $z = 90 \ cm = \frac{90}{100} = 0.9 \ m$

The second spring's spring constant is $k_2 = 350 \ N/m$

We denote the distance from the center of the box to the left edge as x.

At equilibrium,

The force exerted by the first spring is

$F_1 = k_1 * (0.8 -x)$

while the force from the second spring is

$F_2 = k_2 * [ 0.9 - (0.9 -x)]$

Thus, at equilibrium,

$F_1 = F_2$

Substituting values gives us

$k_1 * (0.8 -x) = k_2 * [ 0.9 - (0.9 -x)]$

which leads to

$200 * (0.8 -x) = 350 * [ 0.9 - (0.9 -x)]$

resulting in

$160 -200x) = 350x$

and finally,

$160 =550x$

this simplifies to

$x =0.291 \ m$

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