Question

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

Answer 1

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:

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

$\Delta s = 189.166 m$