Very Simple Work-Kinetic Energy Theorem Question?

Question

How should the values for the work done by friction obtained by using A) the definition of work and B) the work-kinetic energy theorem relate to one another? Determine the ratio of work done by method A to method B. Report these ratios and describe what they mean.

I just need to know how I would go about doing this.

Answer 1

A)

Energy can be defined as the capacity for doing work. It may exist in a variety of forms and may be transformed from one type of energy to another. However, these energy transformations are constrained by a fundamental principle, the Conservation of Energy principle. One way to state this principle is "Energy can neither be created nor destroyed". Another approach is to say that the total energy of an isolated system remains constant. Work = force x distance

B)

The change in the kinetic energy of an object is equal to the net work done on the object.

KE= (1/2)mv^2

This fact is referred to as the Work-Energy Principle and is often a very useful tool in mechanics problem solving. It is derivable from conservation of energy and the application of the relationships for work and energy, so it is not independent of the conservation laws. It is in fact a specific application of conservation of energy. However, there are so many mechanical problems which are solved efficiently by applying this principle that it merits separate attention as a working principle.

You can see that friction is a force that does work on an object. W = Fd

It also can change the energy of the onbject by slowing it down. Change in KE = Change in[(1/2) mv^2]

Answer 2

By definition work energy WE = driving force F X distance s driven.  For friction, to keep something moving at constant velocity, F = kN = k mg cos(theta); where k is the friction coefficient and N is the normal weight N = mg cos(theta) and theta is the slope of the surface over which the something m is moved.  From the WE = Fs = k mg cos(theta) s; where s is the distance the m mass is moved along the surface at a constant velocity by the force P pushing against the friction force F.  In many problems the motivating push P = mg sin(theta), which is simply the weight (W = mg) of the mass vectored along the surface of the ramp at theta incline.  As P = F, we find that k = sin(theta)/cos(theta) = tan(theta) when the mass m starts to slide.

Total energy of a system is TE = KE + PE + WE; where KE and PE are kinetic and potential energy if any.  To find the WE in this case, we find the total energy TE and subtract the KE and PE from it.  For example, using your friction example, suppose TE = mgH which is the potential energy of a mass m at the top of an incline H high from zero potential level.

We let our mass m slide down the ramp and note its velocity as it slides off the bottom of the ramp.  PE = 0 at this point because the mass is at ground zero.  KE = 1/2 mV^2, where V is the measured velocity.  Therefore, from energy conservation, TE = mgH = KE + WE and WE = mgH - KE = mgH - 1/2 mV^2; where g = 9.81 m/sec^2.

Not clear what your ratio question is leading up to.  If we are talking about the same m and the same surface with the same slope, then the two work energies would be the same as we are talking about the same work against friction.  Thus, WE = TE - KE = Fs; so that 1 = (TE - KE)/WE is the ratio for the KE method over the WE done.