无摩擦力下物体在斜面上的运动——牛顿运动定律的应用及解答

1. Box’s 质量 = 2 公斤， 重力加速度 = 9.8 m / s². Find (a) the net force which accelerates the box downward (b) magnitude of the box’s 促进.

解决方案

已知：

质量 (m) = 2 kg

重力加速度 (g) = 9.8 米/秒²²

重量 （w）= mg = （2）（9.8）= 19.6 牛顿

w_x = w sin 30 = (19.6)(0.5) = 9.8 Newton

w_y = w cos 30 = (19.6)(0.5√3) = 9.8√3 Newton

解决方案：

（一个） 此 net force which accelerates the box

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is w_x.

∑F = w_x

∑F = 9.8 牛顿

（二） magnitude of the acceleration

∑F = ma

9.8 = (2) a

a = 9.8 / 2

a = 4.9 米/秒²

Magnitude of the acceleration is 4.9 m/s², direction of the acceleration is downward.

2. 斜面 is smooth so there is no 摩擦力. Object’s mass is 3 kg, acceleration due to gravity is 9.8 m/s². Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m/s² (c) object is moving upward with a constant acceleration of 2 m/s².

解决方案

已知：

质量 (m) = 3 kg

重力加速度 (g) = 9.8 米/秒²²

重量 (w) = mg = (3)(9.8) = 29.4 牛顿

w_x = w sin 30 = (29.4)(0.5) = 14.7 Newton

w_y = w cos 30 = (29.4)(0.5√3) = 14.7√3 Newton

解决方案：

(a) The magnitude of the force F if an object is at rest

Newton’s first law of motion states that if an object is at rest, the net force acts on the object is zero.

∑F = 0

F – w_x = 0

F = w_x

F = 14.7 牛顿

(b) The magnitude of the force F if an object is moving downward at a constant 2 m/s²

∑F = ma

w_x – F = m a

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7– 6

F = 8.7 牛顿

(c) The magnitude of the force F if an object is moving upward at a constant 2 m/s²

∑F = ma

F – w_x = ma

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 牛顿

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