If a $10 k g$ $10kg$ object moving at $3 \frac{m}{s}$ $3 m/s$ slows down to a halt after moving $\frac{5}{2} m$ $5/2 m$ , what is the friction coefficient of the surface that the object was moving over?

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If a $10 k g$ $10kg$ object moving at $3 \frac{m}{s}$ $3 m/s$ slows down to a halt after moving $\frac{5}{2} m$ $5/2 m$ , what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2016-03-24T14:39:16

$= 0.18$ $=0.18$

Explanation:

Here work done against force of friction = KE of the moving body
$μ m g \times d = \frac{1}{2} m v^{2}$ $mumgxxd=1/2mv^2$
given
$m = 10 k g$ $m =10kg$
$d = \frac{5}{2} m = 2.5 m$ $d=5/2m=2.5m$
$v = 3 \frac{m}{s}$ $v =3m/s$
$μ = ?$ $mu=?$
$:.mumgxxd=1/2mv^2$
$=>mucancelmgxxd=1/2cancelmv^2$
$=>mu =v^2/(2gd)=3^2/(2xx10xx2.5)=0.18$

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If a $10 k g$ $10kg$ object moving at $3 \frac{m}{s}$ $3 m/s$ slows down to a halt after moving $\frac{5}{2} m$ $5/2 m$ , what is the friction coefficient of the surface that the object was moving over?

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If a $10 k g$ $10kg$ object moving at $3 \frac{m}{s}$ $3 m/s$ slows down to a halt after moving $\frac{5}{2} m$ $5/2 m$ , what is the friction coefficient of the surface that the object was moving over?