第3问能不能帮我画个图理解一下。---**Extracted Content:** 【练 5】新能源电动汽车现在被广泛使用，如图所示，一辆电动汽车停放在水平地面上，满载时整车质量为 1.5 t，在某段平直路面上，该车满载时 90 s 内匀速行驶了 1.8 km，汽车的牵引力做的功是 4.5×10⁶ J。求: (g 取 10 N/kg) (1)汽车行驶的速度; (2)汽车的牵引力; (3)若该车通过一段长 1 000 m，高 120 m 的上坡路段，求此过程重力做的功。 **Chart/Diagram Description:** * **Image:** An image of a modern passenger car. * **Diagram:** A handwritten diagram depicting an inclined plane, likely representing an uphill road section. * It shows a horizontal line segment at the base, a vertical line segment representing height, and a sloped line segment connecting the top of the vertical line to one end of the horizontal line, forming a right-angled triangle. * The vertical side is labeled "120m". * The sloped side is labeled "1000m". This label appears to be pointing to the length along the incline. * Arrows indicate directions: a vertical downward arrow labeled "G=mg" (likely representing gravity), and an arrow pointing along the incline (likely representing motion or force). * There is also a handwritten label "1000m" near the horizontal line segment, which seems inconsistent with the other label on the slope. Based on the problem text (length 1000m, height 120m of an uphill road section), the 1000m likely refers to the length along the incline, and 120m is the vertical height gained. **Other Relevant Text / Handwritten Notes (Likely student's work, not part of the problem statement):** * (1) 驱动机器人前进的动力F的大小。 * (2) 这段时间内动力F所做的功。 * 匀速直线运动 阻力等于牵引力 * 800 x 0.01 = 40N * 阻力 = 动力 = 40N * (2) 根据 W = FS = 40N x 30m = 1200J * v = s/t * 匀速 * (1) 1.8km = 1800m * 速度 v = s/t = 1800m / 90s = 20m/s * (2) 牵引力 4.5 x 10⁶ J * W = Fs -> F = W/s = 4.5 x 10⁶ J / 1800m = 2.5 x 10³ N * 质量为 1.5t = 1500kg * G = mg = 1500kg x 10 N/kg = 15000N (or 1.5 x 10⁴ N) * 重力做的功 = G * h = 15000 N * 120 m = 1.8 x 10⁶ J