The friction coefficient between an athelete's shoes and the ground is 0-90 . Suppose a superman

step 1 In this problem, we have a man running on the ground. The problem asks for two things. First, th

The friction coefficient between an athelete's shoes and the...

Key Concepts

Coefficient of Friction

This concept relates the frictional force resisting motion to the normal force acting on an object. In problems involving motion on surfaces, the coefficient of friction (commonly represented by ?) dictates the maximum possible force that friction can supply before slipping occurs. It is a dimensionless number that, when multiplied by the normal force, gives the maximum frictional force available for acceleration or deceleration.

Friction-Limited Acceleration

When an object is accelerating on a surface, the maximum acceleration is restricted by the frictional force preventable between the object and the surface. According to Newton’s second law, the maximum possible acceleration is given by a = ?g, where g is the acceleration due to gravity. This acceleration limit is key in problems where the propulsive force can be arbitrarily high, but the frictional force is the limiting factor.

Kinematic Equations for Uniform Acceleration

These are equations that relate displacement, initial velocity, acceleration, and time for objects moving with constant acceleration. They are essential for solving problems where an object starts from rest, accelerates over a known distance, or decelerates to a stop. The relevant equations allow one to determine times, distances traveled, and final velocities under constant acceleration or deceleration.

Deceleration with Friction Constraints

Similar to acceleration, the rate at which an object can decelerate is limited by the friction force available. When stopping, the same maximum deceleration of magnitude ?g applies (but opposite in direction). This concept is used to calculate the minimum stopping distance or time from a given velocity, assuming friction is the only force acting to slow the object down.

Transcript

00:02 In this problem, we have a man running on the ground.

00:07 The problem asks for two things.

00:10 First, the time will it take for the man to reach 50 meters starting from rest, assuming that a frictional force is acting on the man's shoes with a coefficient of 0 .90.

00:30 Next, the problem as asked for the time will it take, for the man to reach full stop after completing the 50 hunters.

00:47 This problem involves applying newton's law of motion and kinematic equations.

01:00 To solve for this problem, we first have to find an expression for the man's acceleration.

01:09 We can do that by using the newton's law motion.

01:12 First, let's draw the forces acting on the man's body.

01:18 First, its weight pointed vertically downwards.

01:23 Let's denote it as f sub w.

01:26 Next, the force perpendicular to the surface, which is f sub n, and the friction force opposite its movement, which is f sub k.

01:48 In this case, the type of friction that is acting on the man is kinetic because the man is moving and then the force applied by the man to make him move in this direction.

02:03 We denote it as f sub a or a applied force.

02:12 Now let's write these forces splitting it into vertical and horizontal components.

02:20 First for the vertical components, the summation of these forces in this direction, let's denote as f sub 1 is equal to 0 according to first law, where it states that a body in equilibrium, the net forces, the south net forces is equal to 0.

02:51 In this plane, the forces are normal force and the weight of the man.

03:00 We denoted as negative since it is offset the direction of the normal force.

03:17 The formula for weight is the product of the muscle.

03:23 Of the body and acceleration due to gravity.

03:31 As such, this equation becomes like this.

03:34 Simplifying the normal force, stress equals to the product of mass times the acceleration due to gravity.

03:43 Now, for the horizontal forces, also the summation of the forces on this plane is also equal to zero.

03:59 And the forces involving this plane are the flight force and the friction null force.

04:21 From second love motion, the applied force is just equal to the product of the mass of the body and the acceleration due to its movement.

04:32 This is the value refers what we are looking for first.

04:37 We denoted as a -sabx.

04:39 And the formula for friction is mu sub k multiplied by the normal force.

05:01 Rearranging, we have m multiplied by a sub x is equals to mu sub k multiplied by normal force.

05:15 Since we know the thermal force is just m g, we can rewrite this equation like this, dividing both sides by the mass to simplify, we have acceleration stress equals to new subk multiplied by the isolation due to gravity.

05:46 Since the coefficient of friction is given, we can solve this equation...

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