An object with a mass of $7 k g$ $7 kg$ is hanging from a spring with a constant of $2 \frac{k g}{s^{2}}$ $2 (kg)/s^2$ . If the spring is stretched by $14 m$ $14 m$ , what is the net force on the object?

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Answer 1 · 2017-05-12T07:00:33

The net force is $= 40.6 N$ $=40.6N$ downwards

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Mass of object $m = 7 k g$ $m=7kg$

The spring constant is $k = 2 k g s^{- 2}$ $k=2kgs^-2$

Extension of the spring is $x = 14 m$ $x=14m$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

The net force is $F = 2 \cdot 14 - 7 g$ $F=2*14-7g$

$= 28 - 7 g = - 40.6 N$ $=28-7g=-40.6N$

There is a net force of $40.6 N$ $40.6N$ acting downwards

An object with a mass of 7 kg7kg7 kg is hanging from a spring with a constant of 2 (kg)/s^22kgs22 (kg)/s^2. If the spring is stretched by 14 m14m 14 m, what is the net force on the object?