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?
I can't give a answer because I am very confused.
Well, no need to be confused! Let me help you out with some humor along the way.
a. To find the output work of the jack, we need to remember that work is equal to force multiplied by distance. Since the car is being lifted, we're essentially working against gravity. And let's face it, gravity can be a real drag sometimes. The output work can be calculated using the formula: output work = weight × distance lifted. So, just multiply the weight of the car by the distance it's lifted.
b. Power is the rate at which work is done. And who doesn't love a powerful car jack? To find the power required to lift the car, we can use the formula: power = work ÷ time. Divide the input work by the time it takes to lift the car, and you've got the power!
c. Efficiency is a measure of how well something performs, and we all know how important it is to perform well. To calculate efficiency, we can use the formula: efficiency = (output work ÷ input work) × 100%. Don't worry if the jack isn't efficient enough, we all have our off days!
Now, armed with these formulas, you can calculate the output work, power, and efficiency of that ideal car jack. Just remember, even if the math gets tricky, laughter is the best tool in your bot-t belt!
No problem, I can help you break it down step by step. Let's solve each part one by one.
a. To find the output work of the jack, we need to know the mechanical advantage of the jack. The mechanical advantage (MA) is calculated by dividing the output distance by the input distance. In this case, the input distance is 0.35 meters and the output distance is the distance the car is lifted.
Given the input distance (d1) = 0.35 meters and the output distance (d2) is not given directly, but we can calculate it using the following formula:
Mechanical Advantage (MA) = Output Distance (d2) / Input Distance (d1)
Since MA = 2750 J / Output Work (W2), we have:
2750 J / Output Work (W2) = d2 / d1
Simplifying the above equation gives us:
Output Work (W2) = 2750 J * d1 / d2
Now we can substitute the given values and calculate the output work.
Output Work (W2) = 2750 J * 0.35 m / d2
b. To calculate the power required to lift the car, we can use the formula:
Power (P) = Work (W) / Time (t)
Here, we need to calculate the work done in lifting the car (output work) from part a, and we are given the time it takes to lift the car, which is 8 seconds.
Power (P) = Output Work (W2) / Time (t)
Substitute the given values to find the power required.
c. Efficiency is the ratio of output work to input work, expressed as a percentage.
Efficiency = (Output Work / Input Work) * 100%
Efficiency = (W2 / Input Work) * 100%
Now you can substitute the values you've calculated to find the efficiency of the jack.
No worries! I'll break it down for you step by step.
a. To find the output work of the jack, we can use the formula:
Output work = Input work - Energy lost
In an ideal system, there is no energy lost, so the output work is equal to the input work. Therefore, the output work of the jack is 2750 J.
b. To calculate the power required to lift the car, we can use the formula:
Power = Work / Time
Since we want to lift the car a distance of 0.35 meters in 8 seconds, the work done is the output work calculated in part a (2750 J), and the time is 8 seconds. Substituting these values into the formula, we get:
Power = 2750 J / 8 s = 343.75 watts
Therefore, the power required to lift the car is 343.75 watts.
c. To calculate the efficiency of the jack, we can use the formula:
Efficiency = (Output work / Input work) * 100
From part a, we know the output work is 2750 J, and the input work is also 2750 J. Substituting these values into the formula, we get:
Efficiency = (2750 J / 2750 J) * 100 = 100%
Therefore, the efficiency of the jack is 100%.
I hope this helps clarify the steps to solve the problem! Let me know if there's anything else I can assist you with.
a. An ideal jack: Work output = Work input = 2750 J.
b P = Work/t = 2750J./8s.= 344 J/s = 344 Watts.
c. Efficiency = Po/Pi = 344/344 = 1.0 = 100 %.