A 2.0 kg piece of wood slides on the surface. The curved sides are perfectly smooth, but the rough horizontal bottom is 30 m long and has a kinetic friction coefficient of 0.20 with the wood. The piece of wood starts from rest 4.0 m above the rough bottom. Where will this wood eventually come to rest?

From the initial release until the wood reaches the state of rest, what amount of work is done by friction?

This problem aims to familiarize with the concepts of dynamic motion which are part of classical dynamic physics. To better understand this topic you should be familiar with kinetic energy, kinetic friction, and energy lost due to friction.

The first term we should be familiar with is kinetic energy, which is the energy that the object maintains due to its motion. It is defined as the work needed to accelerate an object of some certain mass from rest to its given velocity. The object sustains this kinetic energy unless its velocity shifts after attaining it during its acceleration.

Another terminology to keep in touch with is kinetic friction which is described as a force acting between rolling surfaces. A body rolling on the surface undergoes a force in the opposing direction of its motion. The amount of force will rely on the coefficient of kinetic friction between the two surfaces.

The Kinetic friction coefficient is denoted by $\mu_k$ and its value is $0.20$.

The Mass of the wood is $m$ and is given by $2.0 \space Kg$.

The Height above the rough bottom is $h$ and its value is $4.0 \space m$.

The Gravitational force is $g$ and is given as $9.8 m/s^2$.

Part a:

First, we will find the distance $d$, from the initial state, where the wood eventually comes to rest.

According to the Law of conservation of Energy,

Initial Energy = Final Energy,

OR,

Gravitational Potential Energy = Friction Energy.

$mgh = \mu_kgdm$

Inserting the values:

$(2.0)(9.8)(4) = (0.2)(9.8)(2.0)d$

Making $d$ the subject:

$d = \dfrac{78.4}{3.92}$

$d = 20 \space m$

Part b:

To find the total amount of work done by friction, we will find the total initial energy that will be the total work friction has done.

The initial energy is Gravitational Potential Energy given by:

$P.E. = mgh$

Inserting the values:

$= (2.0)(9.8)(4.0)$

$= 78.4 \space J$

Numerical Result

The distance where the wood eventually comes to rest is $20 \space m$.

The total amount of work done by friction is $78.4 \space J$.

Example

A piece of the log having the mass $1.0 \space kg$ falls against a surface. The log has entirely smooth curved sides and a rough horizontal bottom that is $35 \space m$ long. The kinetic friction coefficient of the log is $0.15$. The starting point of the log is $3 \space m$ beyond the rough bottom. Find how much work friction has to do to stop the log.

To find the total amount of work done by friction, we will find the total initial energy that will be the total work friction has done.

The total work done by friction is the initial energy, that is Gravitational Potential Energy, and is given by:

$P.E. = mgh$

Inserting the values:

$= (1.0)(9.8)(3.0)$

$P.E.= 29.4 \space J$

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