Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if $μ$ $mu$ is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

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Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if $μ$ $mu$ is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

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Block of mass $m$ $m$ has initial velocity $u$ $u$ having direction along $+ x$ $+x$ axis. The block stops after covering distance $S$ $S$ causing similar extension in the spring of spring constant $K$ $K$ holding the block. if μ is the coefficient of kinetic friction between the block and surface on which it was moving, the distance $S$ $S$ is given by the equation

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Answer 1 · 2017-07-27T20:27:05

I got a different answer than the published result.
Did I make a mistake!

Explanation:

Assuming that initially the spring is in its equilibrium position.

Initial energy of block is its kinetic energy $= \frac{1}{2} m v^{2}$ $=1/2mv^2$

Using Law of Conservation of energy:
When the block stops it initial kinetic energy is converted in to mechanical potential energy of spring which gets stretched by distance $S$ $S$ and remaining energy is spent doing work against force of friction during its movement.

PE of the spring $= \frac{1}{2} K S^{2}$ $=1/2KS^2$

Force of friction $= μ m g$ $=mumg$

Work done against force of friction $= Force \times distance$ $="Force"xx"distance"$
$= μ m g \times S$ $=mumgxxS$

Equating the initial and final energies we have

$\frac{1}{2} m v^{2} = \frac{1}{2} K S^{2} + μ m g S$ $1/2mv^2=1/2KS^2+mumgS$
$\Rightarrow K S^{2} + 2 μ m g S - m v^{2} = 0$ $=>KS^2+2mumgS-mv^2=0$

Solving the quadratic in $S$ $S$ we get

$S = \frac{- 2 μ m g \pm \sqrt{{(2 μ m g)}^{2} - 4 \times K (- m v^{2})}}{2 K}$ $S=(-2mumg+-sqrt((2mumg)^2 -4xxK(-mv^2)))/(2K)$
$S = \frac{1}{K} (- μ m g \pm \sqrt{{(μ m g)}^{2} + K m v^{2}})$ $S=1/K(-mumg+-sqrt((mumg)^2 +Kmv^2))$

Ignoring the $- v e$ $-ve$ root as the movement is in $+ x$ $+x$ direction we get
$S = \frac{1}{K} (\sqrt{{(μ m g)}^{2} + K m v^{2}} - μ m g)$ $S=1/K(sqrt((mumg)^2 +Kmv^2)-mumg)$

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Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if $μ$ $mu$ is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

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Explanation:

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Science

Math

Humanities

... and beyond

Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if μmu is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

1 Answer

Explanation:

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Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if $μ$ $mu$ is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?