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kifflom [539]
3 years ago
6

A block is sliding on a level surface of varying materials, and so its effective coefficient of friction is variable, 0.1t, wher

e t is the time in seconds for how long the block has been sliding. If the block was initially moving at 43.5 m/s, then how far does it move in 5.1 s
Engineering
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

\Delta s = 189.166 m

Explanation:

The physical for the system is based on Work-Energy Theorem and Principle of Energy Conservation. The system decelerates because of friction before coming to rest:

K_{1} = W_{loss,1 \longrightarrow 2}

\frac{1}{2} \cdot m \cdot v^{2} = \mu (t) \cdot m \cdot g \cdot \Delta s

The distance before stopping is isolated from expression presented above:

\Delta s = \frac{ v^{2}}{2 \cdot \mu(t)\cdot g}

Where \mu (t) = 0.1\cdot t and g = 9,807 \frac{m}{s^{2}}.

By replacing all variables, the needed distance is finally found:

\Delta s = \frac{(43.5 \frac{m}{s})^{2}}{2 \cdot [0.1\cdot (5.1 sec)]\cdot (9.807 \frac{m}{s^{2}} )}

\Delta s = 189.166 m

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