If a $1 k g$ $1 kg$ object moving at $25 \frac{m}{s}$ $25 m/s$ slows down to a halt after moving $625 m$ $625 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $1 k g$ $1 kg$ object moving at $25 \frac{m}{s}$ $25 m/s$ slows down to a halt after moving $625 m$ $625 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-02-29T18:30:03

$u_{k} = 0, 038$ $u_k=0,038$

Explanation:

$v_{f}^{2} = v_{i}^{2} - 2 \cdot a \cdot x$ $v_f^2=v_i^2-2*a*x$
${(\frac{v}{2})}^{2} = v^{2} - 2 \cdot a \cdot 625$ $(v/2)^2=v^2-2*a*625$
$\frac{v^{2}}{4} = v^{2} - 2 \cdot a \cdot 625$ $v^2/4=v^2-2*a*625$
$v^{2} = 4 v^{2} - 8 \cdot a \cdot 625$ $v^2=4v^2-8*a*625$
$3 v^{2} = 8 \cdot a \cdot 625$ $3v^2=8*a*625$
$v = 25 \frac{m}{s}$ $v=25 m/s$
$3 \cdot 25^{2} = 8 . a .625$ $3*25^2=8.a.625$
$3*cancel(625)=8*a*cancel(625)$
$a=3/8 m/s^2" acceleration of the object"$
$F_f=u_k*N" the friction force "$
$N=m*g" N:normal force acting contacting surfaces"$
$F_f=m*a" Newton's the second law"$
$u_k*cancel(m)*g=cancel(m)*3/8$
$u_k=3/(8*9,81)$
$u_k=0,038$

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If a $1 k g$ $1 kg$ object moving at $25 \frac{m}{s}$ $25 m/s$ slows down to a halt after moving $625 m$ $625 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer

Explanation:

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If a 1 kg1kg1 kg object moving at 25 m/s25ms25 m/s slows down to a halt after moving 625 m625m625 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

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Explanation:

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If a $1 k g$ $1 kg$ object moving at $25 \frac{m}{s}$ $25 m/s$ slows down to a halt after moving $625 m$ $625 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?