如圖*所示,間距為L、足夠長的固定光滑平行金屬導軌MN、PQ與水平面成θ角,左端M、P之間連接有電流傳感器和阻...
問題詳情:
如圖*所示,間距為L、足夠長的固定光滑平行金屬導軌MN、PQ與水平面成θ角,左端M、P之間連接有電流傳感器和阻值為R的定值電阻。導軌上垂直停放一質量為m、電阻為r的金屬桿ab,且與導軌接觸良好,整個裝置處於磁感應強度方向垂直導軌平面向下、大小為B的勻強磁場中。在t= 0時刻,用一沿MN方向的力斜向上拉金屬桿ab,使之從磁場的左邊界由靜止開始斜向上做直線運動,電流傳感器將通過R的電流i即時採集並輸入電腦,可獲得電流i隨時間t變化的關係圖線,電流傳感器和導軌的電阻及空氣阻力均忽略不計,重力加速度大小為g。
(1)若電流i隨時間t變化的關係如圖乙所示,求t時刻杆ab的速度υ大小;
(2)在(1)問的情況下,請判斷杆ab的運動*質,並求t時刻斜向上拉力的功率P;
(3)若電流i隨時間t變化規律為i=Imsint,則在0~T時間內斜向上拉力對杆ab做的功W。
【回答】
【標準解答】(1)由乙圖可知,t=t1時刻電路中的感應電流為I1,則t時刻,電流為
i=t ··········································································································· ①(1分)
杆ab切割磁感線產生的感應電動勢為
e = BLυ··············································································································· ②(1分)
根據閉合電路歐姆定律有
e = i( R + r ) ······································································································ ③(1分)
由以上三式解得
υ = t ································································································ ④(1分)
(2)由於是常量,所以杆ab是做勻加速直線運動,其加速度大小為
a = = ························································································ ⑤(1分)
設t時刻水平拉力大小為F,根據牛頓第二定律有
F–BiL–mgsinθ=ma ······················································································· ⑥(1分)
又 P=Fυ········································································································· ⑦(1分)
得 P=t2 + [gsinθ + ] t ···························· ⑧(2分)
(3)設位移x=–Acos t,由導數的物理意義可知,有
υ=A · sint ·························································································· ⑨(1分)
由BLυ=i( R + r )及i=Imsint 得
υ=sint ···················································································· ⑩(1分)
可見,杆ab做簡諧運動。
所以振幅A= ········································································ (11)(1分)
在0 ~ T時間內,重力做功為W1 =–mg · 2Asinθ········································· (12)(1分)
t=T時刻,杆ab的速度大小
υ= 0 ·········································································································· (13)(1分)
在0~T時間內,整個迴路產生的焦耳熱為
Q= ()2( R + r ) · T ·············································································· (14)(1分)
安培力對杆ab做的功為
W2 =–Q ······································································································ (15)(1分)
根據動能定理有
W + W1 + W2= – 0 ········································································· (16)(1分)
聯立以上四式解得
W= + Im2( R + r )T ················································· (17)(1分)
【思維點拔】本題的關鍵在於對電流傳感器得到的電流i隨時間t變化的關係圖線的理解,獲取信息,從加速度定義來確定運動*質,利用數學知識位移的導函數是速度函數來確定簡諧運動的振幅,從而來求重力做功,整個迴路產生的焦耳熱要用電流的有效值來算,然後藉助動能定理加以求解。
知識點:專題四 功和能
題型:綜合題