Patients in a wheelchair
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Patients in a wheelchair
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[post_date] => 2025-01-09 07:30:02
[post_date_gmt] => 2025-01-09 12:30:02
[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.
A hospital is considering revamping its rehabilitation center. In an effort to facilitate movement for handicapped individuals, administrators commissioned a project to reconstruct a specific ramp. Preliminary investigations involved three volunteer patients wheeling their way up the ramp individually and providing feedback on the difficulty of the exercise.
Figure 1. A diagram of the ramp under consideration for reconstruction.
The hospital is further interested in obtaining better wheelchairs. The hospital staff is pursuing qualities such as improved balance, weight, torque, and wheel grip. To consolidate this entire project, they requested the patient volunteers to each use a different wheelchair under consideration for widespread hospital use during their trial.
Wheelchair A has a frame composed of aluminum and has thin rubber wheels. On the other hand, wheelchair B is made of plastic and has thicker rubber wheels. Finally, wheelchair C has increased rigidity due to its titanium build and large leather wheels.
Information involving each wheelchair’s mass, type, and coefficient of friction is compiled in Table 1.
Table 1. Compilation of wheelchair data.
The masses of volunteers A, B, and C are 60, 50, and 70kg respectively.
Table 2. Trigonometric function table of angles 30 and 45 degrees.
[post_title] => Patients in a wheelchair
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[question] => Volunteer B makes her way up the ramp on wheelchair B. She pauses and engages her brakes to catch her breath. What is the force of friction exerted on the wheelchair while at rest on the ramp? (Use 10 m/s2 for the acceleration due to gravity)
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[answer] => 4
[description] => Reason for the Correct Answer:
Remember to incorporate the mass of both the patient and the wheelchair in your calculation.
The forces acting on the patient are gravity, the normal force, and static friction.
The normal force is equal to the vertical component of the force of gravity (relative to the inclined section of the ramp). If the wheelchair is not moving, the static friction force must be equal to the horizontal component of the force of gravity.
In order to calculate the magnitude of static friction, you must know the horizontal component of the force of gravity (relative to the inclined section of the ramp). This is given by multiplying the magnitude of mass * g with the sin of the angle.
Fₛ = sin(θ) x mass x g
Substituting given values, Fₛ = sin(30) x (50kg + 20kg) x (10 m/s²) = 350N
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[answers] => Array
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[0] => Array
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[each_answer] => A. 260N
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[1] => Array
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[each_answer] => B. 210N
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[2] => Array
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[each_answer] => C. 150N
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[3] => Array
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[each_answer] => D. 350N
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => Volunteer C is moving up the ramp with increasing velocity. How many forces are acting on the wheelchair?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Try to draw a force diagram.
The patient must be pushing himself up the ramp.
Other forces include gravity, frictional force, and the normal force. The total number of forces is 4.
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[0] => Array
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[each_answer] => A. 1
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[1] => Array
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[each_answer] => B. 2
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[each_answer] => C. 3
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[3] => Array
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[each_answer] => D. 4
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Volunteer B is moving at constant velocity up the ramp. What must be true about the forces on the wheelchair?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
Use ΣF = ma.
The acceleration is given to you. Plug this into F = ma.
If acceleration is 0, then ΣF = 0. So, the net force on the volunteer is equal to 0.
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[answers] => Array
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[0] => Array
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[each_answer] => A. The force of static friction equals the horizontal component of the normal force
)
[1] => Array
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[each_answer] => B. The net force on the volunteer is equal to zero
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[2] => Array
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[each_answer] => C. The force of kinetic friction equals the horizontal component of the normal force
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[3] => Array
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[each_answer] => D. The force of kinetic friction equals the normal force
)
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => Volunteer A is moving at a constant acceleration of a up the ramp. She is pushing herself with a force of Fp. What is the force of friction on the wheelchair?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
Use ΣF = ma
The horizontal force of gravity and the force of friction are both pointing down the ramp. The Fₚ is up the ramp.
In other words, ΣF = Fₚ – Fբ – Fᵍ =ma
Isolate Fբ
-Fբ = ma + Fᵍ – Fₚ
Divide by (-1): Fբ= – ma – Fᵍ + Fₚ
The horizontal component of gravity is sin(30)mg = (.5)mg [see Table 2].
This is tricky! See if you can draw the two 30-60-90 triangles needed to convince yourself that this is true!
Plugging this value in, we get:
Fբ = – ma – (.5)mg + Fₚ
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[answers] => Array
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[0] => Array
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[each_answer] => A. Fբ = -ma + Fₚ – (.87)mg
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[each_answer] => B. Fբ = -ma + Fₚ – (.5)mg
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[2] => Array
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[each_answer] => C. Fբ = ma + Fₚ + (.5)mg
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[each_answer] => D. Fբ = ma + Fₚ + (.87)mg
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => Volunteer A realizes she is moving too quickly up the ramp, and engages her brakes. This causes her to begin skidding up the ramp. Which of the following regarding forces on the wheelchair is true?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
The brakes do not induce any force on their own. They simply make the wheels stop moving.
The wheel’s interaction with the ground (friction) is what causes the force.
The force of friction always opposes the direction of the motion.
Since the volunteer was going up the ramp, the force of friction will be pointing down the ramp.
This is the same direction as the horizontal component of the force of gravity.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The force of the brakes and the force of friction are pointing in opposite directions
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[1] => Array
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[each_answer] => B. The force of friction and horizontal force of gravity are pointing in opposite directions
)
[2] => Array
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[each_answer] => C. The force of the brakes and the force of friction are pointing in the same direction
)
[3] => Array
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[each_answer] => D. The force of friction and horizontal force of gravity are pointing in the same direction
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