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

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

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Answer 1 · 2016-02-21T13:56:00

$u_{k} ≅ 0, 425$ $u_k~=0,425$

Explanation:

$E_{k} = \frac{1}{2} \cdot m \cdot v^{2} The kinetic energy of object$ $E_k=1/2*m*v^2 " The kinetic energy of object"$
$W=F_f* Delta x " Work doing by friction force"$
$F_f=u_k*N " N: normal force to contacting surfaces"$
$"The kinetic energy turns work"$
$1/2*cancel(m)*v^2=u_k.cancel(m)*g*Delta x$
$1/2*5^2=u_k*9,81*3$
$25=u_k*2*3*9,81$
$u_k=25/(2.3.9,81)$
$u_k~=0,425$

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

1 Answer

Explanation:

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Explanation:

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