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Zorluk: Çok zorInclined Planes

A worker must load a 500 lb500\text{ lb} crate into a truck bed that is 4 ft4\text{ ft} off the ground. He can choose between two ramps. Ramp X is 8 ft8\text{ ft} long and operates as an ideal, frictionless machine. Ramp Y is 16 ft16\text{ ft} long, but has a rough surface that reduces its mechanical efficiency to 50%50\%. How do the required effort force and total work input compare when using Ramp Y instead of Ramp X?

  1. Ramp Y requires the exact same effort force but twice the total work.Cevap
  2. B
    Ramp Y requires half the effort force and the exact same total work.
  3. C
    Ramp Y requires half the effort force but twice the total work.
  4. D
    Ramp Y requires twice the effort force and twice the total work.

Cevap

Ramp Y requires the exact same effort force but twice the total work.
Although Ramp Y is twice as long as Ramp X (which ideally cuts the required force in half), its 50% efficiency means half of the worker's effort is wasted fighting friction. These two factors perfectly cancel each other out, resulting in the exact same effort force of 250 lbs. However, because the worker is applying that same 250 lbs of force over twice the distance (16 ft instead of 8 ft), the total work performed is doubled.

Adım Adım Çözüm

1
Calculate the useful work output required to lift the crate.
Workout=Load×Height=500 lb×4 ft=2,000 ft-lb\text{Work}_{\text{out}} = \text{Load} \times \text{Height} = 500\text{ lb} \times 4\text{ ft} = 2,000\text{ ft-lb}.
This baseline value is the minimum energy required to achieve the lift regardless of the ramp used.
2
Evaluate the effort force and work input for Ramp X.
IMAX=8 ft4 ft=2\text{IMA}_{\text{X}} = \frac{8\text{ ft}}{4\text{ ft}} = 2. EffortX=500 lb2=250 lb\text{Effort}_{\text{X}} = \frac{500\text{ lb}}{2} = 250\text{ lb}. Workin X=250 lb×8 ft=2,000 ft-lb\text{Work}_{\text{in X}} = 250\text{ lb} \times 8\text{ ft} = 2,000\text{ ft-lb}.
Because Ramp X is frictionless (100% efficient), work input equals work output, and the force is simply the load divided by the IMA.
3
Calculate the total work input required for Ramp Y.
Efficiency=WorkoutWorkin\text{Efficiency} = \frac{\text{Work}_{\text{out}}}{\text{Work}_{\text{in}}}. So, 0.50=2,000Workin0.50 = \frac{2,000}{\text{Work}_{\text{in}}}, resulting in Workin Y=4,000 ft-lb\text{Work}_{\text{in Y}} = 4,000\text{ ft-lb}.
Friction wastes energy as heat. A 50% efficiency means twice as much energy must be put into the system to get the required useful work out.
4
Calculate the effort force required for Ramp Y.
Workin=Effort×Distance\text{Work}_{\text{in}} = \text{Effort} \times \text{Distance}. 4,000 ft-lb=EffortY×16 ft4,000\text{ ft-lb} = \text{Effort}_{\text{Y}} \times 16\text{ ft}. EffortY=250 lb\text{Effort}_{\text{Y}} = 250\text{ lb}.
Force can be derived by dividing the total actual work input by the total distance the force is applied.
5
Compare the force and work values between the two ramps.
Effort force is identical (250 lb250\text{ lb} for both). Total work for Ramp Y (4,000 ft-lb4,000\text{ ft-lb}) is twice that of Ramp X (2,000 ft-lb2,000\text{ ft-lb}).
This establishes the final correct comparison.

Anahtar Kavram

Actual Mechanical Advantage and Efficiency
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Konular

Inclined Planes

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