Work, Energy and Friction

功、能与摩擦

Example of Energy Balance Calculations

能量平衡计算示例

In the "friction experiment" you collected data about distances and times for the masses involved. This data allows you to make calculations of the energies involved in the motions. One of the fundamental symmetries of nature is conservation of energy: "Energy is neither created nor destroyed.". This means that for any physical experiment, you should be able to "balance the energy checkbook" and account for the energies involved in different parts of the experiment. The experiment starts with an energy balance in the form of the gravitational potential energy of the mass hanging on the string. As the mass falls to the floor, you withdraw all of that balance, paying it out in the form of the kinetic energies of the falling mass and the moving block on the table above. The string attached to the block on the wooden runway does work on it which is force x distance. That work moves the block against the frictional resistance mmg where the weight is W=mg . It also gives it kinetic energy so that it is still moving for a time after the hanging mass hits the floor. Ultimately, friction wins out and stops the sliding block, and at that time you can say that all the work done on the block has been "withdrawn" from your energy account as it is dissipated against the frictional resistance. By balancing the energy checkbook, we can imply the work done against friction and determine a value for the coefficient of friction which acted between the block and the wooden track on which it is sliding.

在‘摩擦实验’中，你收集了有关涉及质量的距离和时间的数据。这些数据允许你计算运动中涉及的能量。自然界的一个基本对称性是能量守恒：‘能量既不会创造也不会消灭’。这意味着对于任何物理实验，你应该能够‘平衡能量账本’，并核算实验中不同部分所涉及的能量。实验开始时，能量平衡以悬挂于绳子上的质量的引力势能形式存在。当质量下落至地面时，你撤回了所有的平衡能量，以下落质量的动能和桌面上移动的块的动能形式支付出去。连接块的绳子对块做功，该功等于力乘以距离。这项工作使块克服摩擦阻力m mg，其中重量是W=mg。它还给予块动能，使其在悬挂质量撞击地面后仍继续移动一段时间。最终，摩擦力占上风，停止了滑动的块，此时你可以说所有作用在块上的功都已从你的能量账本中‘撤回’，因为它们被消耗在摩擦阻力上。通过平衡能量账本，我们可以推断出克服摩擦力所做的功，并确定块与滑动木轨之间摩擦系数的值。

The following is an example analysis from data taken in the laboratory. The detailed analysis has parallel input boxes where you may enter your data for comparison and determine the coefficient of friction from your own data. The data boxes will default to the values for the examples, so you will have to make sure that your values for the measurements are entered.

以下是从实验室获取的数据示例分析。详细分析包含平行的输入框，您可以在此输入数据进行比较，并根据自己的数据确定摩擦系数。数据框将默认显示示例中的值，因此您必须确保输入的测量值是正确的。

Experimenters: Brett Smith and Julie Griffis

The first step in the analysis is to determine your initial energy balance, the potential energy of the hanging mass.

分析的第一步是确定初始的能量平衡，即悬挂质量的势能。

Potential energy = mgh = (0.1502 kg)(9.8 m/s²)(0.3 m) = 0.44 joules.

势能 = mgh = (0.1502 kg)(9.8 m/s²)(0.3 m) = 0.44 焦耳。

Potential energy = mgh = ( kg)(9.8 m/s²)( m) = joules.

势能 = mgh = (kg)(9.8 m/s²)(m) = 焦耳。

中文译文中的待填/计算数值依次对应：1：mh 2：h 3：pe。实际数值以上方原输入框为准。

The "energy account" that we are considering consists of just the hanging mass and the block to which it is attached by the string. We neglect the string because of its tiny mass, so our account consists of the energy of these two objects. Any energy transferred to the earth (the table, the floor, the wooden track on which it is sliding, etc.) is treated as an energy withdrawal from our system. That energy is of course not lost, since energy is never ultimately lost, but just transferred out of our energy account for the experiment. The first withdrawal occurs when the hanging mass crashes into the floor: it had kinetic energy before the collision, and after the collision it has none. The next step is to calculate that withdrawal.

我们考虑的‘能量账户’仅包括悬挂的质量和通过绳子连接的块。我们忽略绳子，因为它的质量非常小，因此我们的账户仅包括这两个物体的能量。任何转移到地球（桌子、地板、滑动的木制轨道等）的能量都被视为从我们的系统中取出的能量。这种能量当然不会丢失，因为能量从未真正丢失，只是从我们的实验能量账户中转移出去了。第一次能量取出发生在悬挂的质量撞到地板时：碰撞前它有动能，碰撞后则没有。下一步是计算这个能量取出。

Since kinetic energy = (1/2)mv², we need the velocity of impact with the floor. We can calculate the average velocity from the distance divided by the time, but the impact velocity is twice that average. This stage of the energy evaluation is as follows:

由于动能 = (1/2)mv²，我们需要知道物体与地板碰撞时的速度。我们可以通过距离除以时间来计算平均速度，但碰撞速度是该平均值的两倍。能量评估的这一阶段如下：

v_average = h/t = 0.30 m/0.6 s = 0.5 m/s

平均速度 v average = h/t = 0.30 m/0.6 s = 0.5 m/s

v_average = h/t = m/ s = m/s

平均速度 v = h/t = m/s = m/s

中文译文中的待填/计算数值依次对应：1：h2 2：t 3：v。实际数值以上方原输入框为准。

v_impact = 2v_average = 1 m/s

冲击速度为 2 倍平均速度，即 1 m/s

v_impact = 2v_average = m/s

Kinetic energy = (1/2)mv² = (1/2)(0.1502 kg)(1 m/s)² = 0.075 J

动能 = (1/2)mv² = (1/2)(0.1502 kg)(1 m/s)² = 0.075 J

Kinetic energy = (1/2)mv² = (1/2)( kg)( m/s)² = J

动能 = (1/2)mv² = (1/2)(kg)(m/s)² = J

中文译文中的待填/计算数值依次对应：1：mh2 2：v3 3：ke。实际数值以上方原输入框为准。

This amount of energy is now withdrawn from the account, and this completes the energy analysis for the hanging mass. It is sitting on the floor with neither potential or kinetic energy remaining. The remainder of the energy must have been used to do work on the wooden block, the other part of our system. The work done on the wooden block is then

这部分能量现在从账户中被取出，这完成了悬挂质量的能量分析。它正站在地板上，既没有势能也没有动能剩余。剩余的能量必须已经被用来对木块做功，这是我们系统的一部分。对木块所做的功则是

W_{on block} = initial system energy balance - KE lost to floor

作用在块上的功等于初始系统能量平衡减去传给地板的动能

W_{on block} = 0.44 J - 0.075 J = 0.365 J

作用在块上的功为 0.365 焦耳。

W_{on block} = J - J = J

作用在块上的功 = J - J = J

中文译文中的待填/计算数值依次对应：1：pe2 2：ke2 3：wb。实际数值以上方原输入框为准。

Since the work done on the block is given by force x distance, and the distance has been measured, the force exerted on the block by the string can be evaluated.

由于作用在块上的功由力乘以距离给出，而距离已测得，因此可以计算出绳子对块施加的力。

W_{on block} = 0.365 J = Force x distance = F x 0.582 m.

作用在块上的功 = 0.365 J = 力 × 距离 = F × 0.582 m.

F = 0.365 J/0.582 m = 0.63 Newtons.

F = 0.365 J/0.582 m = 0.63 牛顿。

F = J/ m = Newtons.

F = J/ m = 牛顿。

中文译文中的待填/计算数值依次对应：1：wb2 2：d 3：f。实际数值以上方原输入框为准。

Since the force holding the surfaces together (the normal force) is just the weight of the block, we are now able to evaluate the coefficient of friction.

由于使两个表面相互作用的力（法向力）仅仅是块的重量，我们现在可以计算摩擦系数。

F = 0.63 N = mmg = m(0.274 kg)(9.8 m/s²) = m2.68 N

F = 0.63 N = m mg = m (0.274 kg)(9.8 m/s²) = m 2.68 N

The coefficient is then

系数随后是

m = F/m_blockg = 0.23

m = F/m block g = 0.23（专名或术语）

碰撞速度为 2 倍平均速度，即 v_impact = 2v_average = m/s

中文译文中的待填/计算数值依次对应：1：v2。实际数值以上方原输入框为准。

m = ( N) /( kg)g =

m = (N)/(kg·g)

中文译文中的待填/计算数值依次对应：1：f2 2：mb2 3：mu。实际数值以上方原输入框为准。

实验者：Brett Smith 和 Julie Griffis

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Apparatus movie

索引装置电影

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