Suppose an ideal car jack has an input work 2750 J. It lifts a car 0.35 meters in 8 seconds.
a. What is the output work of the jack?
b. How much power is required to lift the car?
c. What is the efficiency of the jack?
See 4:55 PM post.
To find the answers to these questions, you can use the following formulas:
a. Output work (Wout) = Force (F) x Distance (d)
b. Power (P) = Work / Time
c. Efficiency (η) = (Wout / Win) x 100
Let's calculate each part step by step:
a. To find the output work of the jack, we need to calculate the force and distance.
First, let's find the force (F) using the formula:
Force (F) = Input Work (Win) / Distance (d)
Substituting the given values: Win = 2750 J and d = 0.35 m
F = 2750 J / 0.35 m
F ≈ 7857.14 N
Now, let's calculate the output work:
Output work (Wout) = Force (F) x Distance (d)
Wout = 7857.14 N x 0.35 m
Wout ≈ 2749.999 J (rounded to 3 decimal places)
Therefore, the output work of the jack is approximately 2750 J.
b. To find the power required to lift the car, we need to use the formula:
Power (P) = Work (W) / Time (t)
Substituting the given values: W = 2750 J and t = 8 s
P = 2750 J / 8 s
P ≈ 343.75 W
Therefore, the power required to lift the car is approximately 343.75 Watts.
c. To find the efficiency of the jack, we need to use the formula:
Efficiency (η) = (Wout / Win) x 100
Substituting the given values: Wout = 2750 J and Win = 2750 J
η = (2750 J / 2750 J) x 100
η = 1 x 100
η = 100%
Therefore, the efficiency of the jack is 100%.