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?

This solution provides the basic algebraic work which needs to be completed in order to solve both parts of this problem regarding the force of friction and kinetic friction. 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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... h in a distance of 120 m. Use the work-kinetic energy theorem (b ... exerted on the car and compare it with the force found in ... (A) Find the potential energy lost by ...

... N (b) How much work is done on the sledge by ... J (c) What is the increase in internal energy in the ...Forces acting in horizontal direction direction Force due to ...

... 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 ...

Calculating Work and Kinetic Energy. ... Use energy methods to calculate the speed of the 6.00-kg block ... Solution: The net force on the system is the downward force...

... a) The net work done is equal to the change of the kinetic energy. ... Therefore, the friction coefficient is µ = 0.04 x + 0.1 (1) The frictional force is f ...