If a $4 k g$ $4 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows down to a halt after moving $800 m$ $800 m$ , what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2016-06-02T10:05:11

$μ_{k} = 0.013 has no unit$ $mu_k=0.013" has no unit"$

$v_f^2=v_i^2-2*a*Delta x$

$v_f=8" "m/s$

$v_i=16" "m/s$

$Delta x=800" "m$

$a:"acceleration of the object"$

$8^2=16^2-2*a*800$

$64=256-a*1600$

$1600*a=256-64$

$1600*a=192$

$a=192/1600" "m/s^2$

$F_f=mu_k*cancel(m)*g=cancel(m)*a$

$mu_k="coefficient of friction"$

$mu_k*g=a$

$mu_k=a/g$

$mu_k=(192/1600)/(9.18)$

$mu_k=192/(1600*9.18)$

$mu_k=192/14688$

$mu_k=0.013$

If a 4 kg4kg4 kg object moving at 16 m/s16ms16 m/s slows down to a halt after moving 800 m800m800 m, what is the friction coefficient of the surface that the object was moving over?