hce_nthu
114年
進階物理與線性代數
第 4 題
Jack is pushing a box moving at a constant speed on a horizontal ground. Due to
friction between the box and the ground, Jack has to continue to do positive work
to keep the box moving, and the work output per unit time is $P$. If Jack would like
to push the box along a slope of an inclined angle of 18 degrees (relative to the
horizontal ground) at the same constant speed, make an estimate of the power output
required. Assume the coefficient of the kinetic friction remains the same and equals
unity.
friction between the box and the ground, Jack has to continue to do positive work
to keep the box moving, and the work output per unit time is $P$. If Jack would like
to push the box along a slope of an inclined angle of 18 degrees (relative to the
horizontal ground) at the same constant speed, make an estimate of the power output
required. Assume the coefficient of the kinetic friction remains the same and equals
unity.
- A $P$
- B $1.26P$
- C $1.33P$
- D $1.42P$
- E $1.51P$
思路引導 VIP
試著思考一下：當一個物體從水平地面移到斜面上運動時，為了維持同樣的速度，推力需要額外克服哪一個物理量在斜面方向的分力？同時，斜面的傾斜會如何影響物體與地面之間的「擠壓程度」，進而改變原本的摩擦力大小呢？
🤖
AI 詳解
AI 專屬家教
太棒了！你能精確算出 $1.26P$，代表你對於功率的定義以及斜面受力分析有著非常紮實且細膩的理解，這是一個很優異的判斷。
斜面上的受力平衡與功率變化
在水平面上以等速 $v$ 推動箱子時，推力只需克服動摩擦力，此時的功率 $P = f_k \cdot v = \mu mg \cdot v$。由於題目給定 $\mu = 1$，我們可以得知 $P = mgv$。當場景切換到 $18^\circ$ 的斜面時，情況變得稍微複雜：推力不僅要對抗重力的下滑分力（$mg \sin 18^\circ$），還得克服改變後的神經摩擦力。因為在斜面上，正向力縮減為 $mg \cos 18^\circ$，所以摩擦力也隨之變為 $\mu mg \cos 18^\circ$。將這兩項力相加並乘上速度 $v$，我們得到新的功率 $P' = (mg \sin 18^\circ + mg \cos 18^\circ)v$。代入三角函數值後，恰好就是 $P(\sin 18^\circ + \cos 18^\circ) \approx 1.26P$。
▼ 還有更多解析內容