# Sliding Friction Example Problem – Inertia and Motion

Friction is a force resistive to the direction of motion. The friction force is proportional to the Normal force perpendicular to the surface between two objects. The proportionality constant is called the coefficient of friction. There are two coefficients of friction where the difference depends on if the object is in motion or at rest. At rest, the coefficient of static friction is used and if the block is in motion, the coefficient of kinetic friction is used.

This example problem will show how to find the coefficient of kinetic friction of a block moving at a constant velocity under a known force. It will also show how to find how long and how far the block travels before stopping.

Example:
A physics student pulls a 100 kg chunk of stone with constant velocity of 0.5 m/s on a horizontal surface with a horizontal force of 200 N. (Physics students are renowned for their strength.) Assume g = 9.8 m/s2.
a) Find the coefficient of kinetic friction
b) If the rope breaks, how long does it take for the stone to come to rest?
c) How far will the stone travel after the rope breaks?

Solution:
This diagram shows the forces at work as the stone is moving. Fr is the friction force, N is the Normal force, mg is the weight of the block and F is the force the student is exerting to move the block.

Choose a coordinate system where horizontal right is the positive x-direction and vertical up is the positive y-direction. The force of friction is Fr and the Normal force is N. The body is in equilibrium since the velocity is constant. This means the total forces acting on the block are equal to zero.

First, the forces in the x-direction.

ΣFx = F – Fr = 0
F = Fr

The friction force is equal to μkN.

F = μkN

Now we need to know the normal force. We get that from the forces in the y-direction.

ΣFy = N – mg = 0
N = mg

Substitute this normal force into the previous equation.

F = μkmg

Solve for μk Plug in the values for the variables. μk = 0.2

Part b) Once the force is removed, how long until the block stops?

Once the rope breaks, the F force the student supplied is gone. The system is no longer in equilibrium. The forces in the x-direction now equal to ma.

ΣFx = -Fr = ma.

ma = -μkN

Solve for a The forces in the y-direction haven’t changed. From before, N = mg. Plug this in for the Normal force. Cancel the m and we’re left with

a = -μkg

Now that we have the acceleration, we can find the time to stop using

v = v0 + at

the velocity when the stone stops is equal to zero.

0 = v0 + at
at = v0  t = 0.26 s

Part c) How far does the stone travel before it stops?

We have the time to stop. Use the formula:

x = v0t + ½at2

x = (0.5 m/s)(0.26 s) + ½(-1.96 m/s2)(0.26)2
x = 0.13 m – 0.07 m
x = 0.06 m = 6 cm

If you’d like more worked example problems involving friction, check out:
Friction Example Problem – Physics Homework Help
Friction Example Problem – Sliding Down An Inclined Plane
Friction Example Problem 2: Coefficient Of Static Friction

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