Question

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:$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{>}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\mathit{F}}_{\mathbf{1}}\mathbf{<}{\mathit{F}}_{\mathbf{2}}\mathbf{}\mathbf{,}$?

Short answer

The plot is suitable for three situations:

a) Plot 2 for ${\mathrm{F}}_1={\mathrm{F}}_2$

b) Plot 3 for ${\mathrm{F}}_1>{\mathrm{F}}_2$

c) Plot 1 for ${\mathrm{F}}_1<{\mathrm{F}}_2$

Step-by-step solution

The given data

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.

Understanding the concept of kinetic energy and acceleration

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)

a) Calculation to justify the plot f1=f2

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.

b) Calculation to justify the plot f1>f2

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.

c) Calculation to justify the plot f1<f2

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.