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

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Answer 1 · 2017-07-11T08:08:06

The force is $= 63.2 N$ $=63.2N$

The coefficient of friction is

$μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

The frictional force is $F_{r}$ $F_r$

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

The net force on the object is $= F$ $=F$

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

$F - μ_{k} N = m a$ $F-mu_kN=ma$

$F = μ_{k} m g + m a = m (μ_{k} g + a)$ $F=mu_kmg+ma=m(mu_kg+a)$

$μ_{k} = 2$ $mu_k=2$

$a = 12 m s^{- 2}$ $a=12ms^-2$

$m = 2 k g$ $m=2kg$

Therefore,

$F = 2 (2 \cdot 9.8 + 12) = 63.2 N$ $F=2(2*9.8+12)=63.2N$

An object with a mass of 2kg2 kg is on a surface with a kinetic friction coefficient of 2 2 . How much force is necessary to accelerate the object horizontally at 12ms212 m/s^2?