If the length of a $17 cm$ spring increases to $61 cm$ when a $4 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-04-01T13:29:44

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See Below.

We know,

The linear expansion of a elastic body is directly proportional to the force exerted on it.

So, $F prop -Deltax$ . [The minus is to show that the body is trying to regain its original shape.]

So, $F= -kDeltax$ [ $k$ is the constant of proportionality and $k != 0$ ]

Now, Subsititute The Values in the relation.

$color(white)(xx)F = -kDeltax$

$rArr mg = -k(61 - 17)$ [Here, $F = mg$ ]

$rArr 4 xx 10 = -kxx44$ [Taking $g = 10$ $m/s^2$ ]

$rArr k = -(cancel4 xx 10)/cancel44_11$

$rArr k = -10/11$

Hence Explained.

If the length of a 17 cm17 cm spring increases to 61 cm61 cm when a 4 kg4 kg weight is hanging from it, what is the spring's constant?