13.4 Work - kinetic energy theorem (application)

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to work - kinetic energy theorem. The questions are categorized in terms of the characterizing features of the subject matter :

• Constant force
• Variable force
• Maximum kinetic energy

Constant force

Problem 1 : A bullet traveling at 100 m/s just pierces a wooden plank of 5 m. What should be the speed (in m/s) of the bullet to pierce a wooden plank of same material, but having a thickness of 10m?

Solution : Final speed and hence final kinetic energy are zero in both cases. From "work- kinetic energy" theorem, initial kinetic energy is equal to work done by the force resisting the motion of bullet. As the material is same, the resisting force is same in either case. If subscript "1" and "2" denote the two cases respectively, then :

For 5 m wood plank :

$$\begin{array}{l}0-\frac{1}{2}m{v_1}^2=-F{x}_1\\ \frac{1}{2}m{100}^2=FX5\end{array}$$

For 10 m wood plank :

$$\begin{array}{l}0-\frac{1}{2}m{v_2}^2=-F{x}_2\\ \frac{1}{2}m{v_2}^2=FX10\end{array}$$

Taking ratio of two equations, we have :

$$\begin{array}{l}\frac{{v_2}^2}{{100}^2}=\frac{Fx10}{Fx5}=2\end{array}$$

$$\begin{array}{l}\Rightarrow {v_2}^2=2x10000=20000\end{array}$$

$$\begin{array}{l}\Rightarrow v_2=141.4m/s\end{array}$$

Problem 2 : A block of 2 kg is attached to one end of a string that passes over a pulley as shown in the figure. The block is pulled by a force of 50 N, applied at the other end of the string. If change in the kinetic energy of the block is 60 J, then find the work done by the tension in the string.

Solution : We need to know tension and the displacement to find the work by tension. We note here that change in kinetic energy is given. Using "work - kinetic energy" theorem, we find work by the net force - not by the tension alone - on the block :

$$\begin{array}{l}W=\Delta K=60J\end{array}$$

In order to find tension and net force on the block, we draw free body diagram as shown in the figure.

$$\begin{array}{l}T=50N\end{array}$$

and

$$\begin{array}{l}\sum F_y=T-mg\\ \sum F_y=50-2x10=30N\end{array}$$

Let the vertical displacement be "y". Then, the work done by the net force is :

$$\begin{array}{l}30y=60\\ y=2m\end{array}$$

Now, we have values of tension and displacement of the block. Hence, work done by the tension in the string :

$$\begin{array}{l}{W}_T=Txy=50x2=100J\end{array}$$

Note that this example illustrates two important aspects of analysis, using "work - kinetic energy" theorem : (i) work equated to change in kinetic energy is work by net force and (ii) if details like the value of tension and displacement are not known, then we need to employ force analysis to find quantities used for the calculation of work by individual force.

Problem 3 : A block of 1 kg, initially at 10 m/s, moves along a straight line on a rough horizontal plane. If its kinetic energy reduces by 80 % in 10 meters, then find coefficient of kinetic friction between block and horizontal surface.

Solution : From the given data, we can find the change in kinetic energy and hence work done by friction (which is the only force in this case). We do not consider weight or normal force as they are perpendicular to the direction of displacement. Here, $K_f=0.2K_i$ . Applying "Work - kinetic energy" theorem,

$$\begin{array}{l}W=K_f-K_i=0.2K_i-K_i=-0.8K_i\end{array}$$

Now,

$$\begin{array}{l}{K}_i=\frac{1}{2}m{v_i}^2=\frac{1}{2}x1x100=50J\end{array}$$

Combining two equations,

$$\begin{array}{l}W=-0.8x50=-40J\end{array}$$

Now, work by the friction is related to friction as :

$$\begin{array}{l}W=-F_{K}r\end{array}$$

and friction is related to normal force as :

$$\begin{array}{l}{F}_K={\mu }_{K}N\end{array}$$

Combining two equations, we have :

$$\begin{array}{l}W=-{\mu }_{K}Nr\\ {\mu }_K=-\frac{W}{Nr}=-\frac{W}{mgr}\\ {\mu }_K=-\frac{-40}{1x10x10}=0.4\end{array}$$

Variable force

Problem 4 : Velocity - time plot of the motion of a particle of 1 kg from t = 2s to t = 6s, is as shown in figure. FInd the work done by all the forces (in Joule) on the particle during this time interval.

Solution : We are required to find work by all the forces, operating on the particle. No information about force is given. However, states of motion at end points can be read from the given plot. The "work - kinetic energy" theorem is independent of intermediate detail. From the plot. we have : $v_i$ = 2 m/s and $v_f$ = 6 m/s. Now, applying theorem for the end conditions, we get the work by all forces, acting on the particle :

$$\begin{array}{l}W=K_f-K_i\end{array}$$

$$\begin{array}{l}W=\frac{1}{2}m({v_f}^2-{v_i}^2)\\ \Rightarrow W=\frac{1}{2}x1x(6^2-2^2)=0.5x(36-4)\\ \Rightarrow W=16J\end{array}$$

Problem 5 : The displacement of a particle of 0.1 kg moving along x-axis is a function of time as x (in meters) = 2 $t^2$ + t. Find the work done (in Joule) by the net force on the particle during time interval t = 0 to t = 5 s.

Solution : Here, displacement is given as a function of time. This enables us to determine speed at given time instants. This, in turn, enables us to determine change in kinetic energy and hence work by the net force on the particle. Now,

$$\begin{array}{l}v=\frac{dx}{dt}=4t+1\end{array}$$

Speed at t = 0,

$$\begin{array}{l}\Rightarrow v_i=1m/s\end{array}$$

Speed at t = 5 s,

$$\begin{array}{l}\Rightarrow v_f=21m/s\end{array}$$

Applying work-kinetic energy theorem, we get work done by the net force as :

$$\begin{array}{l}W=K_f-K_i\end{array}$$

$$\begin{array}{l}W=\frac{1}{2}m({v_f}^2-{v_i}^2)\\ \Rightarrow W=\frac{1}{2}x0.1x({21}^2-1^2)=22J\end{array}$$

Maximum kinetic energy

Problem 6 : A particle of 0.1 kg is at rest when x = 0. A force given by the function F(x) = ( $9-x^2$ ) N, is applied on the particle. If displacement is in meters, then find the maximum kinetic energy (in Joule) of the particle for x>0.

Solution : The given force is a function of displacement. We can use integral form to determine work. This work can, then, be equated to the change in kinetic energies, using "work - kinetic energy" theorem. Once we have the relation of kinetic energy as a function of "x", we can differentiate and apply the condition for maximum kinetic energy. Now,

$$\begin{array}{l}W=\int F\left(x\right)dx=\int (9-x^2)dx\\ \Rightarrow W=9x-\frac{x^3}{3}\end{array}$$

The particle is initially at rest. It means that its initial kinetic energy is zero. Now, applying "work - kinetic energy" theorem, we have :

$$\begin{array}{l}{K}_f-K_i=K_f=K\\ \Rightarrow K=9x-\frac{x^3}{3}\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{dK}{dt}=9-\frac{3x^2}{3}=9-x^2\end{array}$$

For $\frac{dK}{dt}=0$ , we have :

$$\begin{array}{l}\Rightarrow x=-3m\mathrm{or}+3m\end{array}$$

But for x>0, x = 3 m. Second derivative of kinetic enregy should be negative for this value of "x = 3" so that kinetic energy at this point is maximum.

$$\begin{array}{l}\frac{d^{2}K}{d{t}^2}=-2x=-2x3=-6\end{array}$$

Thus, kinetic energy at x = 3 m is maximum for x>0. It is given by :

$$\begin{array}{l}K=9x-\frac{x^3}{3}=9x3-\frac{3^3}{3}=18J\end{array}$$