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6.57 The Force The Work Needed To Stretch The Rubber Band Can Be Calculated As Follows

The document discusses various calculations related to energy, work, and power, including the work done to stretch a rubber band, the work done by a force acting on a particle, and the power required for a specific mass and velocity. It provides equations and examples for calculating work in different scenarios, such as stretching materials and moving particles along curves. Additionally, it includes specific numerical examples to illustrate the application of these concepts.

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21 views3 pages

6.57 The Force The Work Needed To Stretch The Rubber Band Can Be Calculated As Follows

The document discusses various calculations related to energy, work, and power, including the work done to stretch a rubber band, the work done by a force acting on a particle, and the power required for a specific mass and velocity. It provides equations and examples for calculating work in different scenarios, such as stretching materials and moving particles along curves. Additionally, it includes specific numerical examples to illustrate the application of these concepts.

Uploaded by

민솔
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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<Chap.

6 Energy, Work, and Power>


49, 54, 57, 66, 70, 87
6.57
𝐿𝐿0 −𝑥𝑥 𝐿𝐿20
The force 𝐹𝐹 = 𝐹𝐹0 � 𝐿𝐿0
− (𝐿𝐿 2
�
0 +𝑥𝑥)

The work needed to stretch the rubber band can be calculated as follows:
𝑥𝑥 𝑥𝑥
𝐿𝐿0 − 𝑠𝑠 𝐿𝐿20
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = � 𝐹𝐹(𝑠𝑠)𝑑𝑑𝑑𝑑 = � 𝐹𝐹0 � − � 𝑑𝑑𝑑𝑑
0 0 𝐿𝐿0 (𝐿𝐿0 + 𝑠𝑠)2

𝑥𝑥 2 𝐿𝐿20
∴ 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 = 𝐹𝐹0 �𝑥𝑥 − + − 𝐿𝐿0 �
2𝐿𝐿0 𝐿𝐿0 + 𝑥𝑥

6.70
P = m(bv – c) = 54 kg × (4.27 J/kg⋅m × 5.2 m/s − 1.83 W/kg) ≈ 1100.2 W
10000 𝑚𝑚
The time t = 5.2 𝑚𝑚/𝑠𝑠
= 1923.1 s

∴ Work = P * t ≈ 2116 kJ

6.66

A force pointing in the x-direction is given by 𝐹𝐹 = 𝑎𝑎𝑥𝑥 3/2, where 𝑎𝑎 is a constant. The force does 1.86 kJ of
work on an object as the object moves from 𝑥𝑥 = 0 to 𝑥𝑥 = 18.5 m. Find the constant 𝑎𝑎.

Solution

Using the general definition of work:


����⃗
𝑟𝑟2 ⃗
𝑊𝑊 = ∫����⃗
𝑟𝑟
𝐹𝐹 ∙ 𝑑𝑑𝑟𝑟⃗ (Eq 6.11)
1

Since this case is one-dimensional, we can simplify it as follows:


𝑥𝑥
𝑊𝑊𝑥𝑥 = ∫𝑥𝑥 2 𝐹𝐹𝑥𝑥 𝑑𝑑𝑑𝑑 .
1

Now, substituting 𝐹𝐹 = 𝑎𝑎𝑥𝑥 3/2 , with 𝑥𝑥1 = 0 and 𝑥𝑥2 = 18.5, we proceed with the integration.
18.5
2 5 18.5
𝑊𝑊𝑥𝑥 = � 𝑎𝑎𝑥𝑥 3/2 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑥𝑥 2 � ≈ 𝑎𝑎 ∙ 588.83 𝑚𝑚5/2 = 1860 𝑁𝑁 ∙ 𝑚𝑚
0 5 0

∴ 𝑎𝑎 = 3.16 𝑁𝑁/𝑚𝑚3/2
[여기에 입력]
6.87

A particle moves from the origin to the point 𝑥𝑥 = 3 m, 𝑦𝑦 = 6 m along the curve 𝑦𝑦 = 𝑎𝑎𝑥𝑥 2 — 𝑏𝑏𝑏𝑏, where
𝑎𝑎 = 2 m−1 and 𝑏𝑏 = 4. It’s subject to a force 𝑐𝑐𝑐𝑐𝑐𝑐𝚤𝚤̂ + 𝑑𝑑𝚥𝚥̂, where 𝑐𝑐 = 10 N/m2 and 𝑑𝑑 = 15 N. Calculate
the work done by the force.

Solution

Using the general definition of work:


����⃗
𝑟𝑟2
𝑊𝑊 = � 𝐹𝐹⃗ ∙ 𝑑𝑑𝑟𝑟⃗
����⃗
𝑟𝑟1

Where 𝐹𝐹⃗ = 𝑐𝑐𝑐𝑐𝑐𝑐𝚤𝚤̂ + 𝑑𝑑𝚥𝚥̂ and 𝑑𝑑𝑟𝑟⃗ = 𝑑𝑑𝑑𝑑𝚤𝚤̂ + 𝑑𝑑𝑑𝑑𝚥𝚥̂.

So, the work integral can be written as:


����⃗
𝑟𝑟 3 𝑦𝑦(3) 3 6
2 81𝑎𝑎𝑎𝑎 27𝑏𝑏𝑏𝑏
𝑊𝑊 = � 𝐹𝐹⃗ ∙ 𝑑𝑑𝑟𝑟⃗ = � 𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑 + � 𝑑𝑑 𝑑𝑑𝑑𝑑 = � 𝑎𝑎𝑎𝑎𝑥𝑥 3 — 𝑏𝑏𝑏𝑏𝑥𝑥 2 𝑑𝑑𝑑𝑑 + � 𝑑𝑑 𝑑𝑑𝑑𝑑 = − + 6𝑑𝑑
����⃗
𝑟𝑟1 0 𝑦𝑦(0) 0 0 4 3

Substituting 𝑎𝑎 = 2, 𝑏𝑏 = 4, 𝑐𝑐 = 10 and 𝑑𝑑 = 15:

∴ 𝑊𝑊 = 405 − 360 + 90 = 135 𝐽𝐽

[6.49]

Given physical values: 𝜃𝜃 = 30°, 𝐹𝐹 = 200 N, 𝜇𝜇𝑘𝑘 = 0.18, Δ𝑦𝑦 = 1 m

(a) The box has risen 1 m vertically, so the displacement the box with the force is Δ𝑟𝑟 =
1 𝑚𝑚
= 2 𝑚𝑚. Therefore, the work done during this process is
sin 30
𝑊𝑊 = 𝐹𝐹Δ𝑟𝑟 = (200 N)(2 m) = 400 J

(b) The process was constant speed motion. So, we can use the Newton’s second law.
𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐹𝐹 − 𝑚𝑚𝑚𝑚(sin 𝜃𝜃 + 𝜇𝜇𝑘𝑘 cos 𝜃𝜃) = 0.
The component −𝑚𝑚𝑚𝑚 sin 𝜃𝜃 represents the force of gravity acting in the opposite
direction, and −𝑚𝑚𝑚𝑚𝜇𝜇𝑘𝑘 cos 𝜃𝜃 represents the force of friction. Rewriting the above
equation is as follows.
𝐹𝐹 200 𝑁𝑁
𝑚𝑚 = = 2
≈ 31.1 kg
𝑔𝑔(sin 𝜃𝜃 + 𝜇𝜇𝑘𝑘 cos 𝜃𝜃) (9.8 m⁄s )(sin 30 + (0.18) cos 30)

[6.54]

A force 𝐹𝐹⃗ acts in the x-direction with 𝐹𝐹 = 𝑎𝑎𝑥𝑥 2 , 𝑎𝑎 = 5.0 N/m2. Then the work done by this force as it

[여기에 입력]
acts on a particle moving from 𝑥𝑥 = 0 𝑡𝑡𝑡𝑡 𝑥𝑥 = 6.0 m is
𝑥𝑥=6.0 𝑥𝑥=6
1 216 ∗ 5
𝑊𝑊 = � 𝐹𝐹 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑥𝑥 3 � = J = 360 J.
𝑥𝑥=0 3 𝑥𝑥=0 3

[여기에 입력]

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