Lesson 7 of 13
In Progress

The Work-Energy Theorem for a Variable Force

We are now familiar with the concepts of work and kinetic energy to prove the work-energy theorem for a variable force. We confine ourselves to one dimension. The time rate of change of kinetic energy is

dK/dt = d/dt (1/2mv2)

= m (dv/dt)v

= F v (from Newton’s Second Law)

= F dx/dt

Thus

dK = Fdx

Integrating from the initial position (xi) to final position (xf), we have

KfKidK = ∫xfxiFdx where Ki and Kf are the initial and final kinetic energies corresponding to xi and xf

or Kf Ki = xfxiFdx …………………………………………………………………….(8a)

From Eq. (7), it follows that,

Kf Ki = W ………………………………………………………………………………..(8b)

Thus, the WE theorem is useful in a variety of problems, it does not, in general, incorporate the complete dynamical information of Newton’s second law. It is an integral form of Newton’s second law. Newton’s second law is a relation between acceleration and force at any instant of time. In this sense, the temporal (time) information contained in the statement of Newton’s second law is ‘integrated over’ and is not available explicitly. Another observation is that Newton’s second law for two or three dimensions is in vector form. In the scalar form, information with respect to directions contained in Newton’s second law is no present.

Example 6
A block of mass m = 1 kg, moving on a horizontal surface with speed vi = 2 ms-1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range,

Fr = -k/x for 0.1 < x < 2.01 m

= 0 for x < 0.1 m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed vf of the block as it crosses the pitch.

Answer

From Eq. (8a)

Here, note that ln is a symbol for the natural logarithm to the base e and not the logarithm to the base 10[ln X = logeX = 2.303 log10X].