If a $1 k g$ $1 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows down to a halt after moving $10 m$ $10 m$ , what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2016-04-22T17:27:00

$μ = 0, 98$ $mu=0,98$

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$v^{2} = v_{i}^{2} - 2 \cdot a \cdot s$ $v^2=v_i^2-2*a*s$

$where ; v = 8 \frac{m}{s} v_{i} = 16 \frac{m}{s}$ $"where ; " v=8 m/s" "v_i=16m/s$
$s = 10 m area of under graph$ $s=10m" area of under graph"$

$a=acceleration$ $"a=acceleration"$

$8^{2} = 16^{2} - 2 \cdot a \cdot 10$ $8^2=16^2-2*a*10$

$64 = 256 - 20 a$ $64=256-20a$

$20 a = 256 - 64$ $20a=256-64$

$20 a = 192$ $20a=192$

$a = \frac{192}{20}$ $a=192/20$

$a = 9, 6 \frac{m}{s^{2}} acceleration$ $a=9,6 m/s^2" acceleration"$

$F = m \cdot a$ $F=m*a$

$F_{f} = μ \cdot N N = m g friction force$ $F_f=mu*N" "N=mg" friction force"$

$mu*cancel(m)*g=cancel(m)*a$

$mu=a/g$

$g=9,81 m/s^2$

$mu=(9,6)/(9,81)$

$mu=0,98$

If a 1 kg1kg1 kg object moving at 16 m/s16ms16 m/s slows down to a halt after moving 10 m10m10 m, what is the friction coefficient of the surface that the object was moving over?