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 $19 m/s^2$ ?

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Answer 1 · 2016-11-17T22:09:45

$584.4 N$

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The forces are marked on the diagram;

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

Applying NSL vertically downwards:

$6g - R = 0$
$:. R = 6g$
$:. R = 58.8N$

Applying NSL horizontally:

$D - F = (6)(19)$
$:. D = 114 + F$
$:. D = 114 + muR$
$:. D = 114 + (8)(58.8)$
$:. D = 114 + 470.4$
$:. D = 584.4 N$

An object with a mass of 6 kg6 kg is on a surface with a kinetic friction coefficient of 8 8 . How much force is necessary to accelerate the object horizontally at 19 m/s^2 19 m/s^2?