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 $1 \frac{m}{s^{2}}$ $1 m/s^2$ ?

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Answer 1 · 2018-06-08T12:45:54

The force is $= 1776 N$ $=1776N$

The acceleration is $a = 1 m s^{- 2}$ $a=1ms^-2$

The mass of the object is $m = 12 k g$ $m=12kg$

Apply Newton's Second Law to calculate the force

$F - F_{r} = m \cdot a = 12 \cdot 1 = 12 N$ $F-F_r=m*a=12*1=12N$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal force (reaction) is $N = m g = 12 \cdot 9.8 = 117.6 N$ $N=mg=12*9.8=117.6N$

The coefficient of kinetic friction is $μ_{k} = 15$ $mu_k=15$

The frictional force is

$F_{r} = N μ_{k} = 117.6 \cdot 15 = = 1764$ $F_r=Nmu_k=117.6*15==1764$

The force is

$F = 12 + F_{r} = 12 + 1764 = 1776 N$ $F=12+F_r=12+1764=1776N$

An object with a mass of 12kg12 kg is on a surface with a kinetic friction coefficient of 15 15 . How much force is necessary to accelerate the object horizontally at 1ms2 1 m/s^2?