# An object with a mass of 3 kg is hanging from a spring with a constant of 8 (kg)/s^2. If the spring is stretched by 2 m, what is the net force on the object?

Aug 16, 2017

$13.43$ $\text{N}$ downwards

#### Explanation:

For this problem, let's consider upwards to be the positive direction.

Gravity is acting downwards on the object, i.e. in the negative direction. So $g$ will be equal to $- 9.81$ ${\text{m s}}^{- 2}$.

Also, the spring is stretched by $2$ $\text{m}$ (also downwards). So it will be equal to a displacement of $- 2$ $\text{m}$.

Now, the net force will be the sum of the upwards and downwards forces acting on the object.

Let's use $F = m a$ and $F = - k x$:

$R i g h t a r r o w {F}_{\text{net}} = m a + \left(- k x\right)$

$R i g h t a r r o w {F}_{\text{net}} = 3$ $\text{kg} \cdot \left(- 9.81\right)$ ${\text{m s}}^{- 2} - 8$ ${\text{kg s}}^{- 2} \cdot \left(- 2\right)$ $\text{m}$

$R i g h t a r r o w {F}_{\text{net}} = - 29.43$ ${\text{kg m s}}^{- 2} + 16$ ${\text{kg m s}}^{- 2}$

$R i g h t a r r o w {F}_{\text{net}} = - 13.43$ ${\text{kg m s}}^{- 2}$

$\therefore {F}_{\text{net}} = - 13.43$ $\text{N}$

Therefore, the net force acting on the object is $13.43$ $\text{N}$ downwards.