An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $2$ $2$ . How much force is necessary to accelerate the object horizontally at $17 \frac{m}{s^{2}}$ $17 m/s^2$ ?

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Answer 1 · 2017-04-08T15:28:55

439.2 $N$ $N$

M=12 Kg
k=2 units
a =17 $m s^{- 2}$ $ms^(-2)$

$F_{K f} = 2 \cdot 12 \cdot 9.8 N$ $F_(Kf)=2*12*9.8 N$
$\Rightarrow$ $=>$ 235.2 N

But we know ,

$F_{N e t} = F_{a p p l i e d} - F_{K f}$ $F_(Net)=F_(applied)-F_(Kf)$

$M \cdot A_{N e t} + F_{K f} = F_{A p p l i e d}$ $M*A_(Net)+F_(Kf)=F_(Applied)$
$\Rightarrow$ $=>$ 439.2 N

An object with a mass of 12kg12 kg is on a surface with a kinetic friction coefficient of 2 2 . How much force is necessary to accelerate the object horizontally at 17ms2 17 m/s^2?