Work - kinetic energy theorem Work - kinetic energy theorem is a generalized description of motion - not specific to any force type like gravity or friction. We shall, here, formally write work - kinetic energy theorem considering an external force. The application of a constant external force results in the change in kinetic energy of the particle. For the time being, we consider "constant" external force. At the end of this module, we shall extend the concept to variable force as well. Let v i be the initial speed of the particle, when we start observing motion. Now, the acceleration of the particle is :

A force moves the block on a horizontal surface Force does work on the block. a = F m Let the final velocity of the particle be v f . Then using equation of motion, v f 2 = v i 2 + 2 a r , v f 2 - v i 2 = 2 ( F m ) r Multiplying each term by 1/2 m, we have : 1 2 m v f 2 - 1 2 m v i 2 = 1 2 m x 2 ( F m ) r = F r ⇒ K f - K i = W This is the equation, which is known as work - kinetic energy theorem. In words, change in kinetic energy resulting from application of external force(s) is equal to the work done by the force(s). Equivalently, work done by the force(s) in displacing a particle is equal to change in the kinetic energy of the particle. The above work - kinetic energy equation can be rearranged as : ⇒ K f = K i + W In this form, work - kinetic energy theorem states that kinetic energy changes by the amount of work done on the particle. We know that work can be either positive or negative. Hence, positive work results increase in the kinetic energy and negative work results decrease in the kinetic energy by the amount of work done on the particle. It is emphasized here for clarity that "work" in the theorem refers to work by "net" force - not individual force.

Work - kinetic energy theorem with multiple forces Extension of work - kinetic energy theorem to multiple forces is simple. We can either determine net force of all external forces acting on the particle, compute work by the net force and then apply work - kinetic energy theorem. This approach requires that we consider free body diagram of the particle in the context of a coordinate system to find the net force on it. F N = F 1 + F 2 + F 3 + ........ + F n Work - kinetic energy theorem is written for the net force as : ⇒ K f - K i = F N . r where F N is the net force on the particle. Alternatively, we can determine work done by individual forces for the displacement involved and then sum them to equate with the change in kinetic energy. Most favour this second approach as it does not involve vector consideration with a coordinate system. ⇒ K f - K i = ∑ W i

Application of Work - kinetic energy theorem Work - kinetic energy theorem is not an alternative to other techniques available for analyzing motion. What we want to mean here is that it provides a specific technique to analyze motion, including situations where details of motion are not available. The analysis typically does not involve intermediate details in certain circumstances. One such instance is illustrated here. In order to illustrate the application of "work - kinetic energy" theory, we shall work with an example of a block being raised along an incline. We do not have information about the nature of motion - whether it is raised along the incline slowly or with constant speed or with varying speed. We also do not know - whether the applied external force was constant or varying. But, we know the end conditions that the block was stationary at the beginning of the motion and at the end of motion. So, ⇒ K f - K i = 0 It means that work done by the forces on the block should sum up to zero (according to "work - kinetic energy" theorem). If we know other forces and hence work done by them, we are in position to know the work done by the "unknown" force. We should know that application of work-kinetic energy theorem is not limited to cases where initial and final velocities are zero or equal, but can be applied also to situations where velocities are not equal. We shall discuss these applications with references to specific forces like gravity and spring force in separate modules.

Example Problem : A block of 2 kg is pulled up along a smooth incline of length 10 m and height 5 m by applying an external force. At the end of incline, the block is released to slide down to the bottom. Find (i) work done by the external force and (ii) kinetic energy of the particle at the end of round trip. (consider, g = 10 m / s 2 ). Solution : This question is structured to bring out finer points about the "work - kinetic energy" theorem. There are three forces on the block while going up : (i) weight of the block, mg, and (ii) normal force, N, applied by the block and (iii) external force, F. On the other hand, there are only two forces while going down. The force diagram of the forces is shown here for upward motion of the block.

A block on an incline Known forces acting on the block (i) work done by external force ( W F ) during round trip The most striking aspect about the external force, "F", is that we do not know either about its magnitude or about its direction. We have represented the external force on the force diagram with an arbitrary vector. Further, external force, F, acts only during up journey. Note that the block is simply released at the end of upward journey. It means that we need to find work by external force during upward journey only. W F = W F(up) + W F(down) = W F(up) + 0 = W F(up) The kinetic energy in the beginning and at the end of "up" motion are zero. Note the wordings of the problem that emphasizes this. From "work - kinetic energy" theorem, we can coclude that sum of the work done by the three forces is equal to zero during upward motion. It is, therefore, clear that if we know the work done by the other two forces, then we shall find out the work done by the external force, "F", as required. Work done by the forces normal to the incline is zero. It follows then that we do not need to consider normal forces. According to "work-kinetic energy" theorem, the sum of work done by gravity and external force for motion up the incline is zero : W G(up) + W F(up) = 0 Thus, we need to compute work done by the gravity in order to compute work by the external force "F". Now, the component of weight parallel to incline is directed downward. It means that it (gravity) does negative work on the block while going up. The component of gravity along incline is "mg sinθ", acting downward. Work done by gravity during up journey is : W G(up) = F r = - m g sin θ X L = - m g L sin θ W G(up) = - 2 x 10 x 10 x ( 5 10 ) = - 100 J Hence, work done by the external force, "F", is : W F = W F(up) = - W G(up) = - ( - 100 ) = 100 J (ii) Kinetic energy at the end of round trip : Initial kinetic energy of the block is zero. The kinetic energy is increased by the work done by the forces acting on the block. According to work - kinetic energy theorem : K f - K i = K f - 0 = W (round-trip) Total work done during round trip by external force is 100 J as computed earlier. Total work done during round trip by gravity is : W G(roundtrip) = - m g L sin θ + m g L sin θ = 0 Hence, total work done during round trip is : W roundtrip = W F(roundtrip) + W G(roundtrip) = 100 + 0 = 100 J ⇒ K f = 100 J For understanding purpose, we again emphasize that work done during up motion is zero as block is stationary in the beginning and at the end during motion up the incline. Net work is done in downward motion only by the gravity, whereupon kinetic energy of the block increases. Finally, we should note that this example was specially designed to illustrate calculation of work by unknown force indirectly, using "work-kinetic energy" theorem. This application, however, depends on the specific situation. For example, had the incline been rough, we would be required to consider friction force as well. This friction, in turn, would depend on the external force. As such, we would not be in a position to calculate work indirectly as in this case. Clearly, in that situation, we would need to know the external force as well to apply "work-kinetic energy" theorem.