# An object with a mass of 7 kg is hanging from a spring with a constant of 3 (kg)/s^2. If the spring is stretched by  5 m, what is the net force on the object?

##### 1 Answer
Mar 28, 2017

The net force on the object is $= 53.6 N$

#### Explanation:

Let's resolve in the vertical direction ${\uparrow}^{+}$

mass, $m = 7 k g$

spring constant, $k = 3 k g {s}^{-} 2$

extension of the spring, $x = 5 m$

${F}_{1} = k x = 3 \cdot 5 = 15 N$

${F}_{2} = - 7 g = - 68.6 N$

The net force is

$F = {F}_{1} + {F}_{2} = 15 - 68.6 = - 53.6 N$

The force is acting downwards.