91Ó°ÊÓ

The given data can be listed below,

• The mass of the wood is \(m = 2\;{\rm{kg}}\)
• The length of the rough surface is,\(d = 30\;{\rm{m}}\)
• The initial height of the wood on rough surface is,\(y = 4.0\;{\rm{m}}\)
• The coefficient of kinetic friction is,\({\mu _k} = 0.20\)

Frictional force produces negative energy as work. Reduced kinetic energy is the result of it acting in the opposite direction of the force.

The kinetic energy of wood will be zero as it returns at the point it started from.so the work energy equation energy is given by,

\({U_{grav,1}} + {F_{other}} = {U_{grav,2}}\)

Here,\({U_{grav,1}}\)is the potential energy at the initial position,\({F_{other}}\)is the other force acting on the wood, and\({U_{grav,2}}\)is the final potential energy of the wood.

The initial potential energy of the wood is given by,

\({U_{grav,1}} = mgy\)

Here, m is the mass of the wood, gis the acceleration due to gravity and yis the initial height of the wood.

Substitute all the values in the above,

\(\begin{aligned}{}{U_{grav,1}} = \left( {2\;{\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.8}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {{\rm{4}}\;{\rm{m}}} \right)\\ = 78.4\;{\rm{J}}\end{aligned}\)

The final gravitational energy of the wood is zero because the wood won't reach the opposite curving side because the friction force's work over the entire length of the rough surface is more than the original potential energy.

So, the distance covered by the wood before coming to rest is given by,

\(\begin{aligned}{}{U_{grav,1}} + {F_{other}} = 0\\{U_{grav,1}} - {\mu _k}mgs = 0\\s = \frac{{{U_{grav,1}}}}{{{\mu _k}mg}}\end{aligned}\)

Here,\({\mu _k}\) is the coefficient of kinetic friction, m is the mass of the wood, gis the acceleration due to gravity and yis the initial height of the wood.

Substitute all the values in the above,

\(\begin{aligned}{}s &= \frac{{{U_{grav,1}}}}{{{\mu _k}mg}}\\ &= \frac{{78.4\;{\rm{J}}}}{{\left( {0.20} \right)\left( {2\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}\left( {\frac{{1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}}}{{1\;{\rm{J}}}}} \right)\\ &= \frac{{78.4}}{{3.92}}\;{\rm{m}}\\ &= 20\;{\rm{m}}\end{aligned}\)

Thus, the distance covered by wood before returning to rest is \(20\;{\rm{m}}\)

The work done by the friction can be given by the work-energy theorem,

\({U_{grav,1}} + {F_{other}} = 0\)

Here,\({U_{grav,1}}\)is the potential energy at the initial position.

Substitute the value of gravitational energy in the above,

\(\begin{aligned}{}78.4\;{\rm{J}} + {F_{other}} &= 0\\{F_{other}} &= - 78.4\;{\rm{J}}\end{aligned}\)

Thus the work done by the friction on the wood is \( - 78.4\;{\rm{J}}\).