A ramp in an administrative office
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- A ramp in an administrative office
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A ramp in an administrative office
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[post_date] => 2025-01-09 07:27:46
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[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.
Slips and falls cause a multitude of injuries that require many millions of dollars in medical treatment every year. According to one international index, the approximated cost of workplace slips, trips, and falls is estimated to be $6.6 billion. In an effort to improve workplace conditions and reduce the amount of these occurrences, one company is investigating ways to reduce falls from workers when walking up a linoleum-floored ramp in the work hallways. A diagram of the ramp is given in figure 1.
Figure 1. The ramp in the administrative office.
One way to prevent these slips and falls is to use improved engineering strategies in designing steps, ramps, and other potential problem areas.
Employee volunteers at this office were provided socks that have a rubber coating on the bottom to improve traction. The coefficient of kinetic friction between the rubber bottomed socks and the linoleum-floored ramps are as follows:
Coefficient of static friction μs=0.5 Coefficient of kinetic friction μk=0.3
[post_title] => A ramp in an administrative office
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[question] => An employee is walking up the hospital ramp. If the employee is walking with normal walking motion, what force most directly prevents the walking individual from slipping?
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[answer] => 3
[description] => Reason for the Correct Answer:
In order to have kinetic friction the two surfaces must be sliding across each other.
To prevent a slip, the person must not be slipping yet.
Since static friction prevents slipping, it is static friction that prevents someone from starting to slip on a ramp.
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[answers] => Array
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[each_answer] => A. Kinetic friction
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[each_answer] => B. Gravitational force
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[each_answer] => C. Static friction
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[3] => Array
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[each_answer] => D. Inertia
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => A male and female volunteer are both standing on the ramp. The man has twice the mass of the woman. Which one is more likely to slip?
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[answer] => 2
[description] => Reason for the Correct Answer:
Use ΣF=ma for a mass sitting on an incline at rest.
The formula for the static frictional force is Fₛ = μₛFₙ
The component of gravity acting along the ramp is mgsinθ, and the normal force on the ramp is given by mgcosθ.
Plugging into ΣF = 0 we can find that mgsinθ – μₛmgcosθ = 0. So you can divide both sides by m, and find that since the mass cancels from the calculation entirely, the mass doesn’t matter. In other words, they are equally likely to slip.)
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[each_answer] => A. The man is more likely to slip since his gravitational force is larger
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[each_answer] => B. They are equally likely to slip
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[each_answer] => C. The woman is more likely to slip since her frictional force is smaller
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[each_answer] => D. It cannot be determined without the exact proportion of the mass of both individuals.
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Consider an employee of mass 70kg whose socks have lost traction with the ramp. He is slipping down the office ramp while in the standing position, what is the acceleration of the individual down the ramp?
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[answer] => 4
[description] => Reason for the Correct Answer:
Use Newton’s Second Law ΣF=ma
The component of gravity acting along the ramp is mgsinθ = (70kg)(9.8m/s²) x sin(20°)
The force of kinetic friction is given by μkmgcosθ=(0.3)(70kg)(9.8m/s²) x cos(20°)
Plugging into ΣF=ma we find that (70kg)(9.8m/s²) x sin(20°) – (0.3)(70kg)(9.8m/s²) x cos(20°) = (70kg)a, so a= (9.8m/s²) x sin(20°) – (0.3)(9.8/s2) x cos(20°)
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[answers] => Array
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[0] => Array
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[each_answer] => A. (9.8m/s²) x sin(20°) – (0.5)(9.8m/s²) x sin(20°)
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[1] => Array
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[each_answer] => B. (9.8m/s²) x sin(20°) – (0.5)(9.8m/s²) x cos(20°)
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[2] => Array
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[each_answer] => C. (9.8m/s²) x cos(20°) – (0.3)(9.8m/s²) x cos(20°)
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[each_answer] => D. (9.8m/s²) x sin(20°) – (0.3)(9.8m/s²) x cos(20°)
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[quiz_unique_key] => 2261298308
[question] => What would be the largest angle that this ramp could be increased to while still allowing people to stand on it without slipping?
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[answer] => 2
[description] => Reason for the Correct Answer:
Use Newton’s Second Law for an object in equilibirum ΣF=0
The component of gravity acting along the ramp is mgsinθ
The force of static friction is given by μsmgcosθ.
Since mgsinθ – μₛmgcosθ = 0, it can be written as: mgsinθ = μₛmgcosθ.
Remember that sinθ/cosθ = tanθ, which means that tanθ = μₛ , so tan⁻¹(μₛ ) = θ, so θ = tan⁻¹(0.5)
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[answers] => Array
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[0] => Array
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[each_answer] => A. tan(0.3)
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[1] => Array
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[each_answer] => B. tan⁻¹(0.5)
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[2] => Array
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[each_answer] => C. tan⁻¹(0.3)
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[3] => Array
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[each_answer] => D. tan(0.5)
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => If you were to make a new ramp that had a larger angle, how would the normal force exerted on objects on the new ramp compare to the normal force exerted on objects on the old ramp?
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[answer] => 1
[description] => Reason for the Correct Answer:
Draw a force diagram
The component of gravity acting into the surface of the ramp is mgcosθ)
As you increase angle mg does not change, but cosθ gets smaller, so the normal force will get smaller
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The normal force would be smaller on the new ramp since the component of gravity directed into the surface of the ramp will be smaller
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[each_answer] => B. The normal force would be larger on the new ramp since increasing the angle decreases the magnitude of the force of gravity
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[each_answer] => C. The normal force would be smaller on the new ramp since increasing the angle decreases the magnitude of the force of gravity
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[each_answer] => D. The normal force would be smaller on the new ramp since the component of gravity directed into the surface of the ramp will be larger
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