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

Nov 9, 2016

$41 \text{N}$ downwards.

#### Explanation:

The gravitational force acting on the object is

W = mg = (5 "kg") * (10 "m/s"^2) = 50 "N"

in the downwards direction.

The spring force is given by Hooke's Law,

${F}_{\text{Spring" = kx = (3 "kg/s"^2) * (3 "m") = 9 "N}}$

in the upwards direction.

Therefore, the net force is the sum of all forces acting on the object.

${F}_{\text{Net" = W - F_"Spring}}$

$= 50 \text{N" - 9 "N}$

$= 41 \text{N}$

in the downwards direction.