If a $11 k g$ $11 kg$ object moving at $45 \frac{m}{s}$ $45 m/s$ slows down to a halt after moving $900 m$ $900 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $11 k g$ $11 kg$ object moving at $45 \frac{m}{s}$ $45 m/s$ slows down to a halt after moving $900 m$ $900 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-04-23T07:02:24

By conservation of mechanical energy
Work done against Frictional force = KE of the object
$μ_{k} \times m \times g \times d = \frac{1}{2} \times m v^{2}$ $mu_kxxmxxgxxd=1/2xxmv^2$
$μ_{k} = \frac{v^{2}}{2 g d} = \frac{45^{2}}{2 \times 10 \times 900} = 0.1125$ $mu_k=v^2/(2gd)=45^2/(2xx10xx900)=0.1125$

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If a $11 k g$ $11 kg$ object moving at $45 \frac{m}{s}$ $45 m/s$ slows down to a halt after moving $900 m$ $900 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a 11 kg11kg11 kg object moving at 45 m/s45ms45 m/s slows down to a halt after moving 900 m900m900 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $11 k g$ $11 kg$ object moving at $45 \frac{m}{s}$ $45 m/s$ slows down to a halt after moving $900 m$ $900 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?