An object with a mass of $6 k g$ $6 kg$ is on a surface with a kinetic friction coefficient of $8$ $8$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

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An object with a mass of $6 k g$ $6 kg$ is on a surface with a kinetic friction coefficient of $8$ $8$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

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Answer 1 · 2016-02-20T22:06:40

$F = 524, 88 N$ $F=524,88N$

Explanation:

$F_{f} = u_{k} \cdot m \cdot g Friction Force$ $F_f=u_k*m*g" Friction Force"$
$F_{n} = F - F_{f} Net Force$ $F_n=F-F_f" Net Force"$
$F_{n} = m \cdot a$ $F_n=m*a$
$F - F_{f} = m \cdot a$ $F-F_f=m*a$
$F - u_{k} . m \cdot g = m \cdot a$ $F-u_k.m*g=m*a$
$u_{k} = 8; a = 9 \frac{m}{s^{2}}; m = 6 k g$ $u_k=8 ; a=9 m/s^2 ; m=6 kg$
$F - 8 \cdot 6 \cdot 9, 81 = 6 \cdot 9$ $F-8*6*9,81=6*9$
$F = 54 + 8 \cdot 6 \cdot 9, 81$ $F=54+8*6*9,81$
$F = 524, 88 N$ $F=524,88N$

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An object with a mass of $6 k g$ $6 kg$ is on a surface with a kinetic friction coefficient of $8$ $8$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?

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An object with a mass of $6 k g$ $6 kg$ is on a surface with a kinetic friction coefficient of $8$ $8$ . How much force is necessary to accelerate the object horizontally at $9 \frac{m}{s^{2}}$ $9 m/s^2$ ?