If a $4 kg$ object moving at $5/2 m/s$ slows to a halt after moving $12 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2018-04-15T06:29:53

The coefficient of kinetic friction is $=0.027$

Apply the equation of motion

$v^2=u^2+2as$

To calculate the acceleration

The initial velocity is $u=5/2ms^-1$

The final velocity is $v=0ms^-1$

The distance is $s=12m$

The acceleration is

$a=(v^2-u^2)/(2s)=(0-(5/2)^2)/(2*12)=-0.26ms^-2$

According to Newton's Second Law, the force of friction is

$F_r=ma=4*0.26=1.042N$

The normal force is

$N=mg=4*9.8=39.2N$

The coefficient of kinetic friction is

$mu_k=F_r/N=1.042/39.2=0.027$

If a 4 kg4 kg object moving at 5/2 m/s5/2 m/s slows to a halt after moving 12 m12 m, what is the coefficient of kinetic friction of the surface that the object was moving over?