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An athlete whose mass is 90.0 $\mathrm{kg}$ is performing weight lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 490 $\mathrm{N} .$ He lifts the barbell a distance of 0.60 $\mathrm{m}$ in 1.6 $\mathrm{s}$ . (a) Draw a clearly labeled free-body force diagram for the barbell and for the athlete. (b) Use the diagrams in part (a) and Newton's laws to find the total force that the ground exerts on the athlete's feet as he lifts the barbell.

See Solution.

Physics 101 Mechanics

Chapter 4

Newton's Laws of Motion

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Simon Fraser University

Hope College

McMaster University

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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all right, This problem, you're asked to consider first free body diagram. Excuse me? Have a person in a barbeau and was to force exerted by the ground onto the athlete. So I've drawn the weight lifter and a drawn the barb. Ellis is a brick because the barbells just like an object he moves that 0.6 meters, 1.6 seconds. That's a velocity of 0.375 meters per second. So we have 21st. I want to draw two for by diagrams. Don't be the person one. And on the person one, we have their weight, which we could calculate. It's just, uh, 90 times before everybody ground onto the athlete. Okay, so we have, um, their weight, which we calculate to start, and that is going to be 90 times 9.8. So but my calculator, the Indy Times 9.8 you don't need to There's with Okay. Yes, we have 882 Nunes. Now it's going up on the person is the normal force because even equal ops reaction. So the normal force from the ground was athlete. And that's what I need to solve before we consult that we need to consider the free by diagram of the weight. All right, well, the wait or I should say Bar Bell saying, Wait twice in this kind of confusing. So the bar? Well, yes. Better. Okay, Barbeau. It has a downward wait of 400 Newtons that's given to us. All right. Now we need to think about the Upward Force and the Net force on the objects that goes up Sunnis. Be some acceleration. So we know using her equations. You know that the final velocity squared west initial squared is equal to two times acceleration on the object times, its distance minus original position. Its final velocity was 0.375 start at zero. Own explorations in the X directions would say extra time with the vertical execute. These wise would be more appropriate. Yeah, they were appropriately is wise. They were using vertical direction. We should be more careful and try to use labels that represent the axes were working with. All right, so we're trying to solve for the solution, so that's still unknown. It moved a total of 0.6 meters and we can say that started out. Uh, zero has been from 0.6 meters, where zero is a starting position. All right, is he will 1.2 times acceleration. The direction that is be equal to whatever 0.375 square it is. I think I typed in the wrong one, second 0.14 So the exploration of why is able to 0.14 divided by 1.2 0.117 or 0.1 to a mystic using two decimal places. Well, I could say used. Yeah, they used three desperately is 8.375 So I'm just gonna use three here. 30.117 is with that rounds to is get around. Um, since acceleration, the Barbeau has we know that it's 490 Newtons will be divide that by 9.8 we can get its mass is equal to so for 90 divided by 9.8 is 50 kilograms nice and even. All right, so that means that the we know that the force needed to be If we take 50 and we multiply that times 0.117 get 5 25 So the net upward force needed to be 5.85 Newtons, but remember that it has a downer force of 4 90 So if our net upward force was 5 25 Newtons, we needed to overcome the weight of the barbell which was working a downward direction with a force great enough that I when I overcome with that, I get this net amount of 5.5 and so that's pretty straightforward. You could calculate it. We could no, to Ah X, minus 4 90 needs to equal 5.85 And then we would sat foot on their side and we get that The upward force had to be 495 point 85 Nunes, which would give us the desired net force. And we did all this work because now we know they are free. Body diagram is almost complete, but now we can finish the person free by diagram. It's the person. If I push upward with the force of four or 95.85 Newtons, that means there has to be a force in the opposite direction and equal opposite force, which would hit which would push on the person and therefore pushed the person into the ground with that same force. So 4 95.8 five Newtons. So the person is at rest. They're not falling through the ground when this forces exerted on them. Therefore, the normal force from the ground onto the person has to equal this some. So we know that normal force is equal to do A two plus 495.85 13 soups So many commas 1377 0.85 Nunes So what is the told forces exerted by the ground on Fadhli? Um what if circled here and white or, in other words, this number here and then Our free by diagrams are as do are as shown here in the red and the blue and we are done.

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