(1)
a = ug = 5m/s²
v² - vo² = 2ax x=0.9m
解得v = 5m/s
Ek(A) = mv²/2 = 25/2 J
Ep(D) = mg*2R = 8J
Ek(D) = 9/2 J = mv'²/2 -->v' = 3m/s
F向 = mg + FN = mv'²/R = 45/2 N
FN = 25/2 N
(2)
t2 = √(2h/g) = 0.4s
x = v'*t2 = 1.2m
(3)
对传送带进行受力分析,它受到物体对它向左的摩擦力,还有电动机对它向右的力.
t1 = (5-4)/5 = 0.2s
X物=0.2 *(5+4)/2 = 0.9m
X传=0.2 * 6 = 1.2m
ΔX = 0.3m
f=umg=5N
W=f*ΔX=1.5 J
电动机速度不变.由动能定理得0 = W(电) - W
即W(电) = 1.5 J 此为电动机做功的一部分.
因物体动能的变化最终原因也是电动机.
则ΔEk=9/2 J
故总电动机做功为W(电) + ΔEk = 6 J