Lang Harmon
08/31/2024 · Junior High School

PHYS2310/2310H Worksheet Work and kinetic energy-I 3. The total work required to increase the speed of a certain car from rest to \( 2.0 \mathrm{~m} / \mathrm{s} \) is 1500 J . What is the mass of the car?

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为了求解这个问题,我们需要使用动能定理。动能定理表明,一个物体所受的总功等于其动能的变化。具体步骤如下: ### 步骤 1:确定已知量和未知量 - 已知量: - 初速度 \( u = 0 \mathrm{~m/s} \) - 末速度 \( v = 2.0 \mathrm{~m/s} \) - 总功 \( W = 1500 \mathrm{~J} \) - 未知量: - 质量 \( m \) ### 步骤 2:写出动能定理公式 动能定理公式为: \[ W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \] ### 步骤 3:代入已知量 由于初速度 \( u = 0 \mathrm{~m/s} \),所以公式可以简化为: \[ W = \frac{1}{2} m v^2 \] 将已知量代入公式: \[ 1500 = \frac{1}{2} m (2.0)^2 \] ### 步骤 4:解方程求质量 首先计算 \( (2.0)^2 \): \[ (2.0)^2 = 4.0 \] 然后将其代入方程: \[ 1500 = \frac{1}{2} m \cdot 4.0 \] 简化方程: \[ 1500 = 2.0 m \] 最后,解方程求 \( m \): \[ m = \frac{1500}{2.0} \] \[ m = 750 \mathrm{~kg} \] ### 结论 因此,这辆汽车的质量是 \( 750 \mathrm{~kg} \)。

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汽车的质量是 \( 750 \mathrm{~kg} \)。
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