An object with a mass of $5 k g$ $5 kg$ is on a surface with a kinetic friction coefficient of $3$ $3$ . 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 · 2017-07-11T12:47:16

The force is $= 182 N$ $=182N$

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 $= F$ $=F$

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

$F - μ_{k} N = m a$ $F-mu_kN=ma$

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

$μ_{k} = 3$ $mu_k=3$

$a = 7 m s^{- 2}$ $a=7ms^-2$

$m = 5 k g$ $m=5kg$

Therefore,

$F = 5 (3 \cdot 9.8 + 7) = 182 N$ $F=5(3*9.8+7)=182N$

An object with a mass of 5kg5 kg is on a surface with a kinetic friction coefficient of 3 3 . How much force is necessary to accelerate the object horizontally at 7ms27 m/s^2?