An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $15$ $15$ . How much force is necessary to accelerate the object horizontally at $4 \frac{m}{s^{2}}$ $4 m/s^2$ ?

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Answer 1 · 2017-04-07T07:16:05

The force is $= 1812 N$ $=1812N$

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The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

The normal force is $N = m g$ $N=mg$

So,

$F_{r} = μ_{k} \cdot m g$ $F_r=mu_k*mg$

Resolving in the horizontal direction $\to^{+}$ $rarr^+$

$F - F_{r} = m a$ $F-F_r=ma$

So,

$F = F_{r} + m a$ $F=F_r+ma$

$F = μ_{k} m g + m a$ $F=mu_kmg+ma$

$= m (μ_{k} g + a)$ $=m(mu_kg+a)$

$= 12 (15 \cdot 9.8 + 4)$ $=12(15*9.8+4)$

$= 12 \cdot 151 = 1812 N$ $=12*151=1812N$

An object with a mass of 12 kg12kg12 kg is on a surface with a kinetic friction coefficient of 15 15 15 . How much force is necessary to accelerate the object horizontally at 4 m/s^24ms2 4 m/s^2?