Figure 7-16a shows two horizontal forces that act on a block that is sliding to the right across a frictionless floor. Figure F-16b shows three plots of the block’s kinetic energy K versus time t. Which of the plots best corresponds to the following three situations:$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{>}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{<}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}$?

a) Two horizontal forces ${\mathrm{F}}_1$and ${\mathrm{F}}_2$are acting on the block that is sliding to the right.

b) Three plots of the block’s kinetic energy versus the time graph are given.

Using the concept of kinetic energy with the speed, we can get that the velocity square is linked to a particle's acceleration. This acceleration in turn is linked to the force due to Newton's send law of motion and hence, the acceleration behavior determined by the net force due to the horizontal forces will accordingly relate the force behavior to the energy plots.

Formula:

The force due to Newton’s second law,

F =ma (1)

For ${\mathrm{F}}_1={\mathrm{F}}_2$

As, ${\mathrm{F}}_1$and ${\mathrm{F}}_2$have the same magnitude but opposite direction, the net force on the block must be zero.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_1-{\mathrm{F}}_2 \\ =0 \end{gathered}$$

As net force is zero, it means kinetic energy is conserved.

Hence, plot 2 represents that kinetic energy is constant even though time passes.

For${\mathrm{F}}_1>{\mathrm{F}}_2$

So, the net force on the block is given as:

${\mathrm{F}}_{\mathrm{net}=-}{\mathrm{F}}_1+{\mathrm{F}}_2$

So, the net force is negative. That’s why acceleration is also negative according to equation (1).

As negative acceleration decreases, kinetic energy also decreases and graph 3 shows that kinetic energy decreases.

Hence, plot 3 represents this case.

For ${\mathrm{F}}_1<{\mathrm{F}}_2$

So, the net force acting on the block is given as:

${\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_2-{\mathrm{F}}_1$

As net force is positive, acceleration is also positive according to equation (1).

As acceleration is positive, velocity increases with respect to time, so kinetic energy also increases.

Hence, plot 1 indicates an increase in kinetic energy.