A 2.0- kg piece of wood slides on the surface. The curved sides are perfectly smooth, but the rough horixontal bottom is 30 m long and has a kinetic friction coefficient of 0.20 with the wood. The iece of wood starts fromrest 4.0 m above the rough bottom (a) Where will this wood eventually come to rest? (b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?

Question

Laith Petty · Accepted Answer

a) Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom  U1=K2  K2=mgy=2(9.8)(4)=78.4 J  Now apply work-energy relation to the motion along the rough bottom  K1+U1+Wother=K2+U2  μkmgs=78.4  , then solve for s  s=78.4μkmg=20 m  b) Friction does -78.4 J of work

Vasquez · Accepted Answer

Given:- Mass of the piece of wood, m=2.0 kg- Length of the rough horizontal bottom, d=30 m- Kinetic friction coefficient between the wood and the surface, μk=0.20- Initial height of the wood, hi=4.0 m(a) To determine where the piece of wood will eventually come to rest, we can use conservation of mechanical energy. The initial mechanical energy of the system is equal to the potential energy at the initial height, mghi, where g is the acceleration due to gravity. When the piece of wood comes to rest, all of the initial mechanical energy will have been dissipated by friction, so we can write:mghi=WfrictionThe work done by friction, Wfriction, can be calculated as the product of the force of friction, Ffriction, and the distance over which it acts, d:Wfriction=FfrictiondThe force of friction can be calculated using the normal force, FN, which is equal to the weight of the piece of wood, mg, where g is the acceleration due to gravity. The force of friction is given by:Ffriction=μkFNSubstituting these equations and solving for the distance the piece of wood travels before coming to rest, drest, we get:mghi=μkmgdrestdrest=hiμk=4.0&amp;nbsp;m0.20=20&amp;nbsp;mTherefore, the piece of wood will eventually come to rest 20 m from the starting point.(b) The total work done by friction can be calculated using the equation for work done by a constant force, which is given by:Wfriction=Ffrictiond=μkFNd=μkmgdSubstituting the given values, we get:Wfriction=0.20×2.0&amp;nbsp;kg×9.81&amp;nbsp;m/s2×30&amp;nbsp;m=117.7&amp;nbsp;JTherefore, the total amount of work done by friction is 117.7 J.

Jeffrey Jordon · Accepted Answer

Answer:(a) 20m(b) 117.7 JExplanation:(a) To determine where the piece of wood will eventually come to rest, we can use the equations of motion for constant acceleration. The only force acting on the wood once it starts sliding is the force of friction, which opposes the motion of the wood. The acceleration of the wood can be calculated using the equation:a=Ffrictionmwhere Ffriction is the force of friction and m is the mass of the wood. The force of friction is given by:Ffriction=μkFNwhere μk is the kinetic friction coefficient and FN is the normal force acting on the wood. The normal force is equal to the weight of the wood, mg, where g is the acceleration due to gravity. Therefore, we can write:Ffriction=μkmgSubstituting this equation into the equation for acceleration, we get:a=μkmgm=μkgSince the wood starts from rest, we can use the equation:d=12at2to calculate the distance the wood travels before coming to rest. Solving for t, we get:t=2da=2dμkgSubstituting the given values, we get:t=2×4.0&amp;nbsp;m0.20×9.81&amp;nbsp;m/s2=2.02&amp;nbsp;sFinally, we can use the equation:drest=12at2to calculate the distance the wood travels before coming to rest. Substituting the calculated values, we get:drest=12μkgt2=12×0.20×9.81&amp;nbsp;m/s2×(2.02&amp;nbsp;s)2=20&amp;nbsp;mTherefore, the piece of wood will eventually come to rest 20 m from the starting point.(b) To calculate the total amount of work done by friction, we can use the equation:Wfriction=∫Ffrictiondxwhere Ffriction is the force of friction and dx is an infinitesimal distance over which the force is applied. Since the force of friction is constant, we can write:Wfriction=Ffriction∫dx=Ffrictiondwhere d is the distance over which the force is applied. Substituting the given values, we get:Wfriction=μkmgd=0.20×2.0&amp;nbsp;kg×9.81&amp;nbsp;m/s2×30&amp;nbsp;m=117.7&amp;nbsp;JTherefore, the total amount of work done by friction is 117.7 J.

A 2.0- kg piece of wood slides on the surface. The curved sides are perfectly smooth, but the rough horixontal bottom is 30 m long and has a kinetic f

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy