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 $7 \frac{m}{s^{2}}$ $7 m/s^2$ ?

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Answer 1 · 2018-02-18T17:45:58

Consider,
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where $μ_{k} = 8.0$ $mu_k = 8.0$ (which is like super glue!).

$SigmaF_y = F_N - mg = 0$
$=>F_N = mg$

$therefore F_N = 6kg * (9.8m)/s^2 = 58.8N$

Hence,

$SigmaF_x = F_a - F_"fr" = ma$

$=> F_a - mu_k*F_N = ma$

$=> F_a = m(a+mu_k*F_N)$

$therefore F_a = 6kg * ((7m)/s^2 + 8.0*58.8N) approx 3kN$

That's some serious work we need to do!

An object with a mass of 6 kg6kg6 kg is on a surface with a kinetic friction coefficient of 8 8 8 . How much force is necessary to accelerate the object horizontally at 7 m/s^27ms2 7 m/s^2?