# An object with a mass of 6 "kg" is on a surface with a kinetic friction coefficient of 15. How much force is necessary to accelerate the object horizontally at 1 "m/s"^2?

Mar 3, 2016

$890 \text{N}$

#### Explanation:

From Newton's Second Law, we know that

$\Sigma \vec{F} = m \cdot \vec{a}$

Now, to know $\Sigma \vec{F}$, ask yourself, what are the forces present?

The weight is balanced with the normal force. All that is left is the applied force and the kinetic friction. The applied force is stronger than the kinetic friction, or else the object would not speed up.

So we write

${F}_{\text{applied" - F_"friction}} = m \cdot a$

To calculate ${F}_{\text{friction}}$, we multiply the normal force (which is equal to the weight of the object) by the coefficient of kinetic friction.

F_"friction" = (15) * (6 "kg") * (9.8 "m/s"^2)

$= 882 \text{N}$

Now, substitute all the values we have.

F_"applied" - 882 "N" = (6 "kg") * (1 "m/s"^2)

${F}_{\text{applied" = 6 "N" + 882 "N}}$

$= 888 \text{N}$

$\approx 890 \text{N}$