An 85 kg cheerleader stands on a scale that reads in kg.

b. What does the scale read if the 85 kg cheerleader lifts the 50 kg cheerleader upward with an acceleration of 2.0 m/s²?

Question

An 85 kg cheerleader stands on a scale that reads in kg.

b. What does the scale read if the 85 kg cheerleader lifts the 50 kg cheerleader upward with an acceleration of 2.0 m/s²?

Accepted Answer

Hey everyone. So this problem is working with Newton's 2nd and 3rd laws. We have a worker of mass 78 kg standing on a digital scale that outputs measurements in kilograms were asked to work out the display on the scale when the worker lifts a 42 kg box upwards accelerating it at three m per second squared Are multiple choice answers. Here are a 120 kg. B 133 kg. C 144 kg or D 90. kg. Alright, so the first thing we're gonna do here is draw a free body diagram for the box by itself. After that, we'll use Newton's second law kind of work out what those forces look like. And then we'll draw the free body diagram for the worker while that worker is lifting the box. Alright, so first the box, we have weight acting in the negative y direction and then we have the force of the worker on the box acting in the positive y direction. So that's the force that it takes to lift that box upwards from Newton's second law, we can recall that the sum of the forces equals mass times acceleration. So for the box, we will write the force of the worker on the box minus the weight equals mass of the box trans acceleration of the box. Wait, we can recall is given by mass times gravity. So we'll add that to the other side. Mass of the box acceleration of the box plus mass of the box times gravity. We have everything we need to solve for the force of the worker on the box. So the maps of the box was given to us in the problem 42 kg, the acceleration of the box was given as 3m/s squared. Again, the weight of the box 42 kg. And we can recall gravity is a constant 9.8 m/s squared. Plug that into our calculator and get the force Of the worker on the box to be newtons. Next, we are going to draw our free body diagram for the worker. So just like the box, we have weight in the negative Y direction, the worker is standing on the scale. So we'll have a normal force and the positive Y direction. And then we also have the force that the box is exerting on the worker. So as the worker is standing there, the box is going to be exerting a downward force on the worker. So we'll call that force of the box on the worker So from Newton's third law, we can recall that we have the forces are equal and opposite. And so we can write that as force of worker on box is equal to negative force of box on worker. We're going to use Newton's second law again here to sum the forces in the y direction. So we'll have some of those forces equal to mass times acceleration. So we've got positive normal force minus weight minus force of will go we've already solved for worker on box. So I'm going to keep that annotation, their worker on box and that is equal to mass of the worker times the acceleration, The worker is just standing there. So we know that the acceleration is zero And therefore that whole term goes to zero. So now let's look at what we have, we can solve for the normal force. We know the weight is given by mass times gravity. We know our force of the worker on the box. So we can solve for the normal force. We can recall that the normal force is mass times gravity when we are standing on like a scale. And so dividing that then by the gravity will give us the weight will give us the mass, sorry. And then the masses, that is what we are looking for in the solution to this problem. So that's kind of where we're going with this. So N equals F worker on box plus weight. Again, that's gonna be mass of now, the worker is holding the box. So that's massive. The worker plus, oh, so sorry. It's just the mass of the worker. And you might think that it's the mass of the worker plus the box. But it's not because we are already taking into account the force of the box on the worker separately. So it is massive, the worker multiplied by correct. So our normal force is going to be newtons plus 78 kg. The mass of the worker given in the problem just pull back up there. Yep, Time 9.8 m/s square. So our normal force while the worker is moving the box Is 11302. Alright. And then the last step here, our normal force is mass times gravity. So 1302 divided by gravity 9.8 m per second squared is going to give the display of the mass on the scale. We'll plug that in and get 133 kg. And so that is the answer to this problem that aligns with choice B. So B is the correct answer. That's all we have for this one. See you in the next video.

An 85 kg cheerleader stands on a scale that reads in kg. b. What does the scale read if the 85 kg cheerleader lifts the 50 kg cheerleader upward with an acceleration of 2.0 m/s²?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Weight and Normal Force

Net Force