If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{80}{g}$ $u_k=80 /g$ , how far will the object continue to move?

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{80}{g}$ $u_k=80 /g$ , how far will the object continue to move?

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Answer 1 · 2017-01-29T17:20:17

The object will travel only $0.10 m$ $0.10m$ (because a kinetic coefficient of $\frac{80}{g}$ $80/g$ is very large!)

Explanation:

This problem is most easily done by conservation of energy.
Two energy forms are involved:
A change in kinetic energy: $\frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ $1/2mv_f^2-1/2mv_i^2$

Frictional heating: $muF_NDelta d$

The equation for conservation of energy is, in this case

$1/2mv_f^2-1/2mv_i^2+mumgDelta d=0$

Notice that we can divide every term by the mass m, (including the right side of the equation)

Also, $v_f=0$ , and inserting the values for $mu$ , $v_i$ and g, we get:

$-1/2(4)^2+(80/g)(g)Delta d=0$

Cancel the two $g$ 's

$-1/2(4)^2+(80)Delta d=0$

$-8+80Delatd=0$

$80Deltad=8$

$Deltad=0.1m$

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{80}{g}$ $u_k=80 /g$ , how far will the object continue to move?

1 Answer

Explanation:

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{80}{g}$ $u_k=80 /g$ , how far will the object continue to move?