In curling, a 19-kg granite stone is released on ice and slides to rest on a target 28.4 m away

In curling, a 19-kg granite stone is released on ice and...

Key Concepts

Friction and Coefficient of Kinetic Friction

Friction is the resistive force that opposes the relative motion of surfaces in contact. The coefficient of kinetic friction is a dimensionless factor that relates the frictional force to the normal force. It is determined by dividing the force of friction by the product of the mass of the object and the acceleration due to gravity, reflecting the interaction between the surfaces.

Kinematics of Constant Acceleration

Kinematics equations for constant acceleration describe the relationship between displacement, initial and final velocities, acceleration, and time. These equations allow one to determine variables such as the time it takes for an object to come to rest when the acceleration (or deceleration) is known.

Newton's Second Law

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This concept is key in relating forces, such as friction, to the resulting motion or deceleration of an object.

Free Body Diagram

A free body diagram is a visual representation that isolates an object and shows all the force vectors acting on it. This tool is crucial for systematically analyzing the forces at play, which in turn facilitates setting up the equations needed to solve problems involving motion and friction.

Transcript

00:04 In this problem, it is given that a granite stone slides over ice and it comes and it travels a distance of 28 .4 meter before coming to rest.

00:18 Now in the problem it is given that initial velocity of the granite stone is 1 .5 meter per second and it travels a distance of 28 .4 meter before coming to rest.

00:36 So let us say at this location the stone comes to rest.

00:43 All right.

00:44 So we have to find, we have to find acceleration of the stone, time required for this stone to stop and the coefficient of kinetic friction of the surface.

00:57 So we have to basically find the value of kinetic friction so we can write mu -k.

01:02 All right.

01:03 So what will be the force diagram for this stone? so there will be a normal region.

01:09 That will act in vertically upward direction, there will be weight of the stone that will act in vertically downward direction and there will be friction force that will act opposite to the relative motion.

01:24 So using this force diagram what we can write that there is no motion in vertical direction therefore in vertical direction forces must be balanced.

01:34 So in vertical direction f net must be equals to zero.

01:48 Therefore n minus m g must be equals to 0 so n will be equals to m g all right we have got one relation for normal force now what will be the equation of motion in horizontal direction so let us say that the ball is moving with acceleration a in forward direction so net force acting on the ball or on the stone in the direction in the direction of acceleration will be minus f so from equation of motion or using newton's second law we can write minus f will be equals to m into a so we can replace f by muo into n so mu is the mu k basically here so mu k into n that is the coefficient of kinetic friction and it will be equals to mass into acceleration now we have just got the value of normal force is m g so we can replace normal force as mu k into m g will be equals to mass into acceleration now in this expression mass gets cancelled so what will be the expression for acceleration so acceleration will be equals to minus mu k into g all right we have got the value of acceleration in terms of mu k let us say this is equation 2 now we will use kinematics of the body it is saying that initially the stone was moving with 1 .5 meter per second and finally it is stopping so final velocity is 0 meter per second and for this it travels a distance of 28 .4 meter so s is 28 .4 meter and acceleration we have just calculated a is minus mu k into g all right so what is the kinematic equation that will relate to these parameters we can use v square is equal to u square plus 2 as so if we put the value of v here so v is 0 so it will be 0 square is equal to u square that is 1 .5 whole square plus 2 into acceleration that is mu k into g into s that is 28 .4 so what will be the value of mu k so mu k will be equal to to 1 .5 square divided by 2 into 28 .4 into g.

04:25 G is 9 .8 meter per second square.

04:29 So this value comes out to be 0 .004...

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