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Sagot :
Answer:
Approximately [tex]0.31\; \rm m[/tex], assuming that [tex]g = 9.81\; \rm N \cdot kg^{-1}[/tex].
Step-by-step explanation:
Initial kinetic energy of the sled and its passenger:
[tex]\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}[/tex] .
Weight of the slide:
[tex]\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}[/tex].
Normal force between the sled and the slope:
[tex]\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}[/tex].
Calculate the kinetic friction between the sled and the slope:
[tex]\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}[/tex].
Assume that the sled and its passenger has reached a height of [tex]h[/tex] meters relative to the base of the slope.
Gain in gravitational potential energy:
[tex]\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}[/tex].
Distance travelled along the slope:
[tex]\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}[/tex].
The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:
[tex]\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}[/tex].
In other words, the sled and its passenger would have lost (approximately) [tex]((137 + 68.1)\, h)\; {\rm J}[/tex] of energy when it is at a height of [tex]h\; {\rm m}[/tex].
The initial amount of energy that the sled and its passenger possessed was [tex]\text{KE} = 63\; {\rm J}[/tex]. All that much energy would have been converted when the sled is at its maximum height. Therefore, when [tex]h\; {\rm m}[/tex] is the maximum height of the sled, the following equation would hold.
[tex]((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}[/tex].
Solve for [tex]h[/tex]:
[tex](137 + 68.1)\, h = 63[/tex].
[tex]\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}[/tex].
Therefore, the maximum height that this sled would reach would be approximately [tex]0.31\; \rm m[/tex].
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