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ICE Princess25 [194]

3 years ago

8

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

ion while momentarily coming to rest. 1)If the initial speed of the box were doubled, how far x2 would the spring compress?

1 answer:

inn [45]3 years ago

7 0

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

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We know that Fick's law is given by,

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Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

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