Explanation of Work Done by A Variable Force

Explanation of Work Done by A Variable Force

A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 6.3 is a plot of a varying force in one dimension.

If the displacement \(\Delta\)x is small, we can take the force F(x) as approximately constant and the work done is then \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % 4vaiabg2da9iaadAeadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH % uoarcaWG4baaaa!3EF4! \Delta W = F\left( x \right)\Delta x\)

This is illustrated in Fig. 6.3(a). Adding successive rectangular areas in Fig. 6.3(a) we get the total work done as \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabgw % KianaaqahabaGaamOramaabmaabaGaamiEaaGaayjkaiaawMcaaiab % gs5aejaadIhaaSqaaiaadIhadaWgaaadbaGaamyAaaqabaaaleaaca % WG4bWaaSbaaWqaaiaadAgaaeqaaaqdcqGHris5aaaa!4434! W \cong \sum\limits_{{x_i}}^{{x_f}} {F\left( x \right)\Delta x} \)

where the summation is from the initial position x_i to the final position x_f.

If the displacements are allowed to approach zero,then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 6.3(b). Then the work done is 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % Gaeyypa0ZaaCbiaeaacqGHuoarcaWG4bGaeyOKH4QaaGimaaWcbeqa % aiGacYgacaGGPbGaaiyBaaaakmaaqahabaGaamOramaabmaabaGaam % iEaaGaayjkaiaawMcaaiabgs5aejaadIhaaSqaaiaadIhadaWgaaad % baGaamyAaaqabaaaleaacaWG4bWaaSbaaWqaaiaadAgaaeqaaaqdcq % GHris5aaGcbaGaeyypa0Zaa8qCaeaacaWGgbWaaeWaaeaacaWG4baa % caGLOaGaayzkaaGaamizaiaadIhaaSqaaiaadIhadaWgaaadbaGaam % yAaaqabaaaleaacaWG4bWaaSbaaWqaaiaadAgaaeqaaaqdcqGHRiI8 % aaaaaa!5924! \begin{array}{l} W = \mathop {\Delta x \to 0}\limits^{\lim } \sum\limits_{{x_i}}^{{x_f}} {F\left( x \right)\Delta x} \\ = \int\limits_{{x_i}}^{{x_f}} {F\left( x \right)dx} \end{array}\)

where ‘lim’ stands for the limit of the sum when \(\Delta\)x   tends to zero.Thus, for a varying force the work done can be expressed as a definite integral of force over displacement (see also Appendix 3.1).

                             fig.6.3(a)                                                             fig.6.3.(b)

(a) The shaded rectangle represents the work done by the varying force F(x), over the small displacement \(\Delta\)x, \(\Delta\)W = F(x) \(\Delta\)x.

(b) adding the areas of all the rectangles we find that for \(\Delta\)x\(\to\) 0, the area under the curve is exactly equal to the work done by F(x).

EXAMPLE 5

A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N.The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate the work done by the two forces over 20 m.

ANSWER 

Fig.6.4. Plot of the force F applied by the woman and the opposing frictional force  f versus displacement.

The plot of the applied force is shown in Fig.

6.4. At x = 20 m, F = 50 N (\(\ne\) 0). We are given that the frictional force f is |f|= 50 N.It opposes motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis.

The work done by the woman is W_F\(\to\) area of the rectangle ABCD + area of the trapezium CEID 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % WaaSbaaSqaaiaadAeaaeqaaOGaeyypa0JaaGymaiaaicdacaaIWaGa % ey41aqRaaGymaiaaicdacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYa % aaamaabmaabaGaaGymaiaaicdacaaIWaGaey4kaSIaaGynaiaaicda % aiaawIcacaGLPaaacqGHxdaTcaaIXaGaaGimaaqaaiabg2da9iaaig % dacaaIWaGaaGimaiaaicdacqGHRaWkcaaI3aGaaGynaiaaicdaaeaa % cqGH9aqpcaaIXaGaaG4naiaaiwdacaaIWaGaamOsaaaaaa!5679! \begin{array}{l} {W_F} = 100 \times 10 + \frac{1}{2}\left( {100 + 50} \right) \times 10\\ = 1000 + 750\\ = 1750J \end{array}\)

The work done by the frictional force is

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vamaaBa % aaleaacaWGMbaabeaakiabgkziUcaa!39E0! {W_f} \to \)area of the rectangle AGHI

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % WaaSbaaSqaaiaadAgaaeqaaOGaeyypa0ZaaeWaaeaacqGHsislcaaI % 1aGaaGimaaGaayjkaiaawMcaaiabgEna0kaaikdacaaIWaaabaGaey % ypa0JaeyOeI0IaaGymaiaaicdacaaIWaGaaGimaiaadQeaaaaa!4627! \begin{array}{l} {W_f} = \left( { - 50} \right) \times 20\\ = - 1000J \end{array}\)

The area on the negative side of the force axis has a negative sign.

The Work-Energy Theorem for Variable Force

The Work-Energy Theorem for 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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiaadsgacaWGlbaabaGaamizaiaadshaaaGaeyypa0ZaaSaaaeaa % caWGKbaabaGaamizaiaadshaaaWaaeWaaeaadaWcaaqaaiaaigdaae % aacaaIYaaaaiaad2gacaWG2bWaaWbaaSqabeaacaaIYaaaaaGccaGL % OaGaayzkaaaabaGaeyypa0JaamyBamaalaaabaGaamizaiaadAhaae % aacaWGKbGaamiDaaaacaWG2baabaGaeyypa0JaamOraiaadAhadaqa % daqaaiaadAgacaWGYbGaam4Baiaad2gacaWGobGaamyzaiaadEhaca % WG0bGaam4Baiaad6gacaGGNaGaam4CaiGacofacaGGLbGaai4yaiaa % d+gacaWGUbGaamizaiaadYeacaWGHbGaam4DaaGaayjkaiaawMcaaa % qaaiabg2da9iaadAeadaWcaaqaaiaadsgacaWG4baabaGaamizaiaa % dshaaaaabaGaamivaiaadIgacaWG1bGaam4CaaqaaiaadsgacaWGlb % Gaeyypa0JaamOraiaadsgacaWG4baaaaa!709C! \begin{array}{l} \frac{{dK}}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{2}m{v^2}} \right)\\ = m\frac{{dv}}{{dt}}v\\ = Fv\left( {fromNewton's{\mathop{\rm Sec}\nolimits} ondLaw} \right)\\ = F\frac{{dx}}{{dt}}\\ Thus\\ dK = Fdx \end{array}\)

Integrating from the initial position (x_i) to final postion (x_f),we have 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WGKbGaam4saaWcbaGaam4samaaBaaameaacaWGMbaabeaaaSqaaiaa % dUeadaWgaaadbaGaamOzaaqabaaaniabgUIiYdGccqGH9aqpdaWdXb % qaaiaadAeacaWGKbGaamiEaaWcbaGaamiEamaaBaaameaacaWGPbaa % beaaaSqaaiaadIhadaWgaaadbaGaamOzaaqabaaaniabgUIiYdaaaa!4855! \int\limits_{{K_f}}^{{K_f}} {dK} = \int\limits_{{x_i}}^{{x_f}} {Fdx} \)

where, K_i and K_f are the initial and final kinetic energies corresponding to x _i  and x _f.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Baiaadk % hacaWGlbWaaSbaaSqaaiaadAgaaeqaaOGaeyOeI0Iaam4samaaBaaa % leaacaWGMbaabeaakiabg2da9maapehabaGaamOraiaadsgacaWG4b % aaleaacaWG4bWaaSbaaWqaaiaadMgaaeqaaaWcbaGaamiEamaaBaaa % meaacaWGMbaabeaaa0Gaey4kIipaaaa!4708! or{K_f} - {K_f} = \int\limits_{{x_i}}^{{x_f}} {Fdx} \)(6.8.a.)

From Eq. (6.7), it follows that

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaWGMbaabeaakiabgkHiTiaadUeadaWgaaWcbaGaamOzaaqa % baGccqGH9aqpcaWGxbaaaa!3CA7! {K_f} - {K_f} = W\)(6.8.b.)

Thus, the WE theorem is proved for a variable force.

While 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. Work-energy theorem involves an integral over an interval 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 whereas the work-energy theorem is in scalar form. In the scalar form, information with respect to directions contained in Newton’s second law is not present.

EXAMPLE 6

A block of mass m = 1 kg,moving on a horizontal surface with speed \(v_i\)=2 m s^–1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force F_r on the block in this range is inversely proportional to x over this range,\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa % aaleaacaWGYbaabeaakiabg2da9maalaaabaGaeyOeI0Iaam4Aaaqa % aiaadIhaaaGaamOzaiaad+gacaWGYbGaaGimaiaac6cacaaIXaGaey % ipaWJaamiEaiabgYda8iaaikdacaGGUaGaaGimaiaaigdacaWGTbaa % aa!47B5! {F_r} = \frac{{ - k}}{x}for0.1 < x < 2.01m\)= 0 for x < 0.1m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed  _\(v\)f of the block as it crosses this patch ?

ANSWER

From Eq. (6.8a)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaWGMbaabeaakiabg2da9iaadUeadaWgaaWcbaGaamyAaaqa % baGccqGHRaWkdaWdXbqaamaalaaabaWaaeWaaeaacqGHsislcaWGRb % aacaGLOaGaayzkaaaabaGaamiEaaaacaWGKbGaamiEaaWcbaGaaGim % aiaac6cacaaIXaaabaGaaGOmaiaac6cacaaIWaGaaGymaaqdcqGHRi % I8aaaa!498F! {K_f} = {K_i} + \int\limits_{0.1}^{2.01} {\frac{{\left( { - k} \right)}}{x}dx} \)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaad2gadaqhaaWcbaGaamyA % aaqaaiaaikdaaaGccqGHsislcaWGRbGaciiBaiaac6gadaqadaqaai % aaikdacaGGUaGaaGimaiaaigdacaGGVaGaaGimaiaac6cacaaIXaaa % caGLOaGaayzkaaaabaGaeyypa0JaaGOmaiabgkHiTiaaicdacaGGUa % GaaGynaiGacYgacaGGUbWaaeWaaeaacaaIYaGaaGimaiaac6cacaaI % XaaacaGLOaGaayzkaaaabaGaeyypa0JaaGOmaiabgkHiTiaaigdaca % GGUaGaaGynaiabg2da9iaaicdacaaI1aGaamOsaaqaaiaadAhadaWg % aaWcbaGaamOzaaqabaGccqGH9aqpdaGcaaqaaiaaikdacaWGlbWaaS % baaSqaaiaadAgaaeqaaOGaai4laiaad2gaaSqabaGccqGH9aqpcaaI % XaGaamyBaiaadohadaahaaWcbeqaaiabgkHiTiaaigdaaaaaaaa!66D2! \begin{array}{l} = \frac{1}{2}m_i^2 - k\ln \left( {2.01/0.1} \right)\\ = 2 - 0.5\ln \left( {20.1} \right)\\ = 2 - 1.5 = 05J\\ {v_f} = \sqrt {2{K_f}/m} = 1m{s^{ - 1}} \end{array}\)

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=log_c X=2.303log₁₀X]