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A skater with an initial speed of 7.60 $\mathrm{m}/\mathrm{s}$ stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

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0.98 $\mathrm{m} / \mathrm{s}^{2}$ 29.5 $\mathrm{m}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

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.

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

02:57

A skater with an initial s…

01:39

06:33

The skier starts from rest…

02:32

During an ice show, a $60.…

01:48

During an ice show, a 60.0…

04:39

A block of ice with mass 2…

04:32

$\bullet$ $\bullet$ A bloc…

01:37

A $\mathrm{A} 68.5 \mathrm…

For this question, you have to use Newton's second law for that. Let me choose the following reference frame vertical axis that I'm calling. Why on a horizontal axis that I'm calling X now for the first item, we apply Newton's second law on the X axis that access you got that the net force is given by the mess off the skater times his acceleration next direction. The net force that is acting on these access is on Lee. That frictional force so minus because it's pointing to the negative side off the X axis. The frictional force Is it close to the mass of the skater times? He's acceleration, but the frictional force we know magnetic situation is given by they connect. Take no coefficients times the normal force. So this is minus the kinetic friction Local official times the normal force which is of course, to m times his acceleration with noticed that the vertical axis there are only two forces acting the weight and the normal and the skater using moving in the vertical axis, and it's not going to move. Therefore, the weight force on the normal force are equal forces, so the normal force in the situation is because of the weight force, then minus you kinetic times. The weight force is increasing the mass off the skater times, his acceleration, the weight forces given by the mass off the skater times, the acceleration of gravity. So miners ethic, frictional coefficient times, am times G is he goes to m time acceleration. We can simplify the masses to get an acceleration. All that's access that is given by minus the kinetic additional coefficient times acceleration of gravity, which you should remember is 9.8 meters per second squared approximately. Then it's his acceleration is minus 0.1 times 9.8, which results in minus 0.98 meters per second squared. The minus sign indicates that the acceleration is pointing to the left, so it's decelerating with a magnitude off 0.98 meters per second squared. Then, on the second item we have to complete, how far will he move before stopping completely? For that, we can use to reach Alice Equation, which tells us that the final velocity squared is equal to the initial velocity squared plus troop times, the acceleration times, the displacement after stopping completely his final velocity will be zero. His initial velocity is 7.6. Then we have. He is acceleration, which is negative. Tu minus trees points 0.19 times a displacement. Now he's served his equation or the displacement. So two times 0.98 times displacement with 7.6 squared, then the displacement is 7.6 squares, divided by true times 0.98 and these results in approximately 29.5 meters. So this is the distance he will travel before stopping completely.

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