Testing new suture material
- Dashboard
- Testing new suture material
Most Recent Score Report
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Previous Score Reports
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Testing new suture material
Array
(
[passage] => WP_Post Object
(
[ID] => 556525
[post_author] => 12815
[post_date] => 2025-01-09 07:28:46
[post_date_gmt] => 2025-01-09 12:28:46
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Researchers are interested in developing a new type of suture for in-combat medical emergencies. They are tasked with developing a strong, flexible, and versatile thread for this purpose. The lead researcher suggests using a hybrid string composed of various materials. This idea is based on the strength of spider webs, which contain many different types of silk. The combined qualities of the silks allow for extraordinary amounts of flexibility and strength in the webs. The researcher proposes that the same might hold true for combat sutures.
The research team develops a thread that combines polyester, silk, and nylon. They then test this polyfilament by developing an apparatus to determine its strength. A box of mass 40 kg hangs from rest from two of these strings as seen in Figure 1. String A is horizontal and connected to the wall. String B is connected to the ceiling at an angle of 30 degrees. For all of the following questions, assume that g = 10 m/s².
Figure 1. The testing apparatus for the strings.
[post_title] => Testing new suture material
[post_excerpt] =>
[post_status] => publish
[comment_status] => closed
[ping_status] => closed
[post_password] =>
[post_name] => testing-new-suture-material
[to_ping] =>
[pinged] =>
[post_modified] => 2025-01-09 07:28:46
[post_modified_gmt] => 2025-01-09 12:28:46
[post_content_filtered] =>
[post_parent] => 0
[guid] => https://medlifemastery.com/?post_type=passage&p=556525
[menu_order] => 0
[post_type] => passage
[post_mime_type] =>
[comment_count] => 0
[filter] => raw
)
[questions] => Array
(
[0] => Array
(
[quiz_unique_key] => 578908434
[question] => What is the force of gravity exerted on the box?
[value] => Array
(
[answer] => 4
[description] => Reason for the Correct Answer:
The fact that there are strings does not affect the calculation for the force of gravity on the box
The force of gravity is another word for the term weight
The force of gravity near Earth is given by Fg = mg
Plugging in the mass of 4.0kg you can find that Fg = 4.0 kg x 10m/s2 = 40 N
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 30 N
)
[1] => Array
(
[each_answer] => B. 20 N
)
[2] => Array
(
[each_answer] => C. 50 N
)
[3] => Array
(
[each_answer] => D. 40 N
)
)
)
[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => What is the vertical component of tension in String B?
[value] => Array
(
[answer] => 4
[description] => Reason for the Correct Answer:
Use ΣF=ma for the vertical forces on the 4.0 kg mass
The box is not accelerating, and so it has acceleration equal to zero
The vertical component of the tension and the force of gravity are the only forces on the box in the vertical direction
Plugging into ΣF=ma for the vertical direction you can find that Tᵧ – mg = 0 or Tᵧ – 40 N = 0, which means that Tᵧ = 40 N
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 30 N
)
[1] => Array
(
[each_answer] => B. 50 N
)
[2] => Array
(
[each_answer] => C. 20 N
)
[3] => Array
(
[each_answer] => D. 40 N
)
)
)
[2] => Array
(
[quiz_unique_key] => 83407773
[question] => What is the magnitude of the Tension in String B?
[value] => Array
(
[answer] => 2
[description] => Reason for the Correct Answer:
Think of the tension in the string as the hypotenuse of a right triangle
There is no acceleration in the vertical direction so try using ΣF=ma in the vertical direction
or the vertical direction ΣF=ma gives Tᵧ – mg = 0 or Tᵧ -40 N
Since sin(30) = Tᵧ/T, T=Tᵧ/sin(30) = 40 N / sin30
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 40 N x sin 30
)
[1] => Array
(
[each_answer] => B. 40 N / sin 30
)
[2] => Array
(
[each_answer] => C. 40 N x cos 30
)
[3] => Array
(
[each_answer] => D. 40 N / cos 30
)
)
)
[3] => Array
(
[quiz_unique_key] => 2261298308
[question] => How does the magnitude of the tension in string B (TB) compare to the magnitude of the tension in string A (TA) ?
[value] => Array
(
[answer] => 3
[description] => Reason for the Correct Answer:
No acceleration means no net force, but it does not mean all force magnitudes are equal
String A does not supply any force in the vertical direction
Since the horizontal components of the tensions must be equal for there to be no horizontal acceleration, and the tension Tʙ also has a vertical component of force, Tʙ must be greater than Tᴀ
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. The magnitude of Tʙ must be smaller than the magnitude of Tᴀ since string A has to hold the box up while being horizontal
)
[1] => Array
(
[each_answer] => B. The magnitude of Tʙ must be smaller than Tᴀ since sin 30 is 0.5 which means the magnitude of Tʙ only has to be half as much as the magnitude of Tᴀ
)
[2] => Array
(
[each_answer] => C. The magnitude of Tʙ must be greater than the magnitude of Tᴀ since their horizontal components are equal but Tʙ also has a vertical component
)
[3] => Array
(
[each_answer] => D. The magnitude of Tʙ must be the same as the magnitude of Tᴀ since no acceleration means no net force
)
)
)
[4] => Array
(
[quiz_unique_key] => 2261298308
[question] => Imagine that you cut String A and carefully lower the box down so that the box hangs at rest only by string B, which is now vertical. Would the tension in string B be greater in this case or smaller compared to when the box was hanging at an angle?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
The force of gravity on the box and the weight of the box remains the same
The vertical component of the force of string B would remain the same
There would be no horizontal component of the force of string B since it hangs vertical, and since the vertical component would remain the same, the magnitude of the tension would decrease
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. The tension in string B would be smaller while hanging vertically since the vertical component would remain the same, but there would be no horizontal component
)
[1] => Array
(
[each_answer] => B. The tension in string B would be the same as the angled case since the force of gravity is the same magnitude as before and is always directed downwards
)
[2] => Array
(
[each_answer] => C. The tension in string B would be larger while hanging vertically since it would now have a larger vertical component of force
)
[3] => Array
(
[each_answer] => D. The tension in string B would be larger while hanging vertically since all the weight of the box is now directed in the vertical direction
)
)
)
)
[total_question] => 5
[correct_answers] => Array
(
[556525|1] => D
[556525|2] => D
[556525|3] => B
[556525|4] => C
[556525|5] => A
)
[hide_display_feedback_settings] =>
[hide_solutions] =>
)
[post_title] => Testing new suture material
[post_excerpt] =>
[post_status] => publish
[comment_status] => closed
[ping_status] => closed
[post_password] =>
[post_name] => testing-new-suture-material
[to_ping] =>
[pinged] =>
[post_modified] => 2025-01-09 07:28:46
[post_modified_gmt] => 2025-01-09 12:28:46
[post_content_filtered] =>
[post_parent] => 0
[guid] => https://medlifemastery.com/?post_type=passage&p=556525
[menu_order] => 0
[post_type] => passage
[post_mime_type] =>
[comment_count] => 0
[filter] => raw
)
[questions] => Array
(
[0] => Array
(
[quiz_unique_key] => 578908434
[question] =>