Question

A hand exerts a constant horizontal force on a block that is free to slide on a frictionless surface (Fig. 6–30). The block starts from rest at point A, and by the time it has traveled a distance d to point B it is traveling with speed $v_{\mathrm{B}}$. When the block has traveled another distance d to point C, will its speed be greater than, less than, or equal to $2v_{\mathrm{B}}$? Explain your reasoning.

FIGURE 6–30 Question 8.

Short answer

The velocity of the block at point C is less than $2v_{\mathrm{B}}$.

Step-by-step solution

Calculation of work done

The value of the work done can be calculated by multiplying the force’s value with the displacement of the object. It is altered linearly to the value of the object’s displacement for a constant value of force.

Apply the work-energy theorem from points A to B

At point A, the speed of the block is \({v_{\rm{A}}} = 0\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\).

According to the work-energy theorem, the work done in displacing the block from point A to point B is given as:

$W_{\mathrm{A}\mathrm{B}}=\frac{1}{2}m{v}_{\mathrm{B}}^2-\frac{1}{2}m{v}_{\mathrm{A}}^2$

\(Fd = \frac{1}{2}mv_{\rm{B}}^2 - \frac{1}{2}m{\left( {0\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}} \right)^2}\) … (i)

Apply the work-energy theorem from points A to C

According to the work-energy theorem, the work done in displacing the block from point A to point C is given as:

$\begin{array}{rl}{W}_{\mathrm{A}\mathrm{C}}& =\frac{1}{2}m{v}_{\mathrm{C}}^2-\frac{1}{2}m{v}_{\mathrm{A}}^2\\ F\times 2d& =\frac{1}{2}m{v}_{\mathrm{C}}^2-\frac{1}{2}m{\left(0\mathrm{m}/\phantom{\mathrm{m}\mathrm{s}}\mathrm{s}\right)}^2\\ 2Fd& =\frac{1}{2}m{v}_{\mathrm{C}}^2\end{array}$

$Fd=\frac{1}{2}m{v}_B^2$ … (ii)

Calculate the velocity at point C

Now, divide equation (ii) by equation (i).

$\begin{array}{rl}\frac{2Fd}{Fd}& =\frac{\frac{1}{2}m{v}_{\mathrm{C}}^2}{\frac{1}{2}m{v}_{\mathrm{B}}^2}\\ v_{\mathrm{C}}^2& =2v_{\mathrm{B}}^2\\ v_{\mathrm{C}}& =\sqrt{2v_{\mathrm{B}}^2}\\ v_{\mathrm{C}}& =\sqrt{2}{v}_{\mathrm{B}}\\ v_{\mathrm{C}}& =\left(1.414\right)v_{\mathrm{B}}\end{array}$

Thus, the velocity of the block at point C is less than $2v_{\mathrm{B}}$.