Work and Energy - Force of Friction and Kinetic Friction

Question: A box of mass m = 20.0 kg is pushed at a constant speed of v = 2.5 m/s, and by a force F = 25.0 N along the horizontal plane. Show the algebraic steps:

a) What is the force of friction between the box and the surface?

b) What is the value of the coefficient of kinetic friction?

Solution Summary

This solution provides the basic algebraic work which needs to be completed in order to solve both parts of this problem. The solution is brief as this is all that is required for the components being asked for in this question.

... same ratio and hence the work done of energy required (in ... of wood is taken as fulcrum, and applying force on the ... doing work but these makes it easy to do work. ...

... Thus the force of friction is Assume the height of the incline is h. Then the ... The work done by friction is By the work-energy theorem, the net work done is ...

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... Hence friction force f = µkN = µk mgcosθ .(1 ... by a displacement by 0.11m, it loses some potential energy. ... this PE is lost as heat due to work done by ...

... of length (5.5) Thus, the work done by the friction force is: (5.6 ... the final mechanical energy is teh initial mechanical energy plus all the work done by ...

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... done by friction is -2.95x105 J; the work done by ... The engine spent energy E 0 . From that Eµ was ... hence the negative sign) and the final energies are potential ...

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