What Is the Minimum Work Needed to Push
Problem 5 Medium Difficulty
(II) What is the minimum work needed to push a 950-kg car 710 m up along a 9.0$^\circ$ incline? Ignore friction.
Answer
$w_{\operatorname{man}}=1.04 \times 10^{6}=1.04 \mathrm{MJ}$
Video Transcript
if this problem were trying to push a car up a hill and to try to find the minimum amount of work that will be required in order to push a car without friction up 710 meters to the top of the hill. I'm gonna draw that angle and the hill is here. The angle is going to be Alfa. Um, Now I'm going to draw the car. The car is going to be a square. Doesn't matter. We know that there is no friction on the car. The next thing I'm gonna do is I'm going to draw a coordinate system based on the center of the car. And so I'm going to draw the X axis here and the Y axes perpendicular to it. Okay, so there we have the Y axes. Let's go ahead and draw the forces that are acting on the car. The three forces acting on the car, neglecting friction. And the 1st 1 is the pusher of the pole push. Could be a pole. Could be a push. Doesn't matter. What matters is the magnitude and the direction of the forced. So I'm going to start with that Forces starting from the center of the car and going up like this. I don't know who was pushing this car of the hill, but I'm going to tell him that it's actually safer to stand up Hill in case the car slides away from you. And so we're going to change that push to a pole. It doesn't change anything else. We're also going to notice that there is a normal force on the car in this problem. It's not going to be a part of the problem, but the normal force is there. And, of course, the mass of the car multiplied by G. The way to the car is here, and I'm going to label that as M G, which is of course, the way to the car. If you look at Alfa and if you look at this angle here, you'll notice that the two of them are complementary, and this angle here and this angle here are also complimentary, which means that my new angle is also equal to Alva. Okay, now I'm going to go back to Black and I'm going to draw one component of M G. M. G is the force that is not in the directions of excellence. So it has two components, one in the UAE direction which must equal the normal force and the other component of mg, which is opposite of the pulling force. So I'm gonna draw that one right here, and I'm going to say that that components is equal to what it's equal to M. G. It's on the opposite side of the angle, so it's equal to M G times the sign of now. Great. Next, let's go ahead and write the formula. The general formula for work work is going to be equal to f times d times that co sign of the angle between them. Now it's important to realize that this data is not the same as the Alfa. That data is the angle between f and deep. They have the same direction here because if you notice what direction he is going up the hill up like this and so they are in the same direction, the angle between them is equal to zero. So the coastline of data in this case is equal to one. I could rewrite the equation of above in terms of the new information we have and I can say that w is equal to P times d. We're not looking for just any amount of work. We're looking for the minimum amount of work required, and that is going to require us to find the minimum amount of pulling force. All right, now let's look att, the pulling force, the pulling force, the minimum pulling your pushing force is going to be equal to just erase that little bit. All right, so there we have it. Okay, so the minimum amount of pulling force is going to happen when F Net is going to be equal to zero. We know that because if there is no force, um, no acceleration up the slope, then we have the minimum amount of pulling force required. When that's true, we know that the force pulling up the hill, which is equal to P minimum, is going to be equal to the force pulling down the hill, which is M g times the sign of Alfa. We're gonna combine our two equations were gonna substitute our value of payment into our equation for W men. And we're gonna find that w men is going to be equal to M G. times the sign of Alfa Times and D so put deep into the equation here. When I do that, I'm done. All I have to do now is substitute the numbers. So what numbers do I have? I know that the mass of the car is going to be equal to 950 kilograms. I know that the distance that he travels or she travels or they travel up the hill is 710 meters and that the angle Eltham is going to be equal to 9.0 degrees. Putting those numbers into this equation, I have that this is equal to 950 kilograms multiplied by 9.81 meters per second squared, which is what my value of G is 9.81 meters per second squared Fergie. Then the distance that we travel is 710 meters and finally the sign of nine degrees. I'm gonna grab a calculators were gonna put make sure that they're in a degree mode, put in the sign of nine degrees multiplied by 710 multiplied by 9.1, multiplied by 950 we end up with a number that looks like this 103512 0.608 And of course, that is going to be jewels because it is meters squared, kilograms per second squared. Um, looking at the problem, I'm going to have to round it down to the proper number of significant figures, and I'm going to give it three significant figures. And so I'm going to say that W men is going to be equal to 1.4 And if you look at the number above its millions one just over one million. So it's 1.4 times 10 to the sixth Jules, which of course, is equal to 1.0 for Mega Jules that the hockey team were that the rugby team is going to have to push the car up the hill 710 meters in order to get out of true
What Is the Minimum Work Needed to Push
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