# An object with a mass of 6 kg is on a surface with a kinetic friction coefficient of  8 . How much force is necessary to accelerate the object horizontally at  19 m/s^2?

Nov 17, 2016

$584.4 N$

#### Explanation:

The forces are marked on the diagram;

$F =$ Friction
$R =$ Reaction from the surface
$D =$ Driving Force (our unknown)

Applying NSL vertically downwards:

$6 g - R = 0$
$\therefore R = 6 g$
$\therefore R = 58.8 N$

Applying NSL horizontally:

$D - F = \left(6\right) \left(19\right)$
$\therefore D = 114 + F$
$\therefore D = 114 + \mu R$
$\therefore D = 114 + \left(8\right) \left(58.8\right)$
$\therefore D = 114 + 470.4$
$\therefore D = 584.4 N$