Question

Is the work required to be done by an external force on an object on a frictionless, horizontal surface to accelerate it from a speed v to a speed 2v (a) equal to the work required to accelerate the object from v=0 to v, (b) twice the work required to accelerate the object from v=0 to v, (c) three times the work required to accelerate the object from v=0 to v, (d) four times the work required to accelerate the object from 0 to v, or (e) not known without knowledge of the acceleration?

Short answer

The correct option is (c), three times the work required to accelerate the object from v=0 to v.

Step-by-step solution

Work-energy principle

This principle implies that the value of work done on a body is always equal to the change in the body’s total kinetic energy.

The change in the body’s total kinetic energy is calculated between the initial and final positions of the body.

The work needed to accelerate the object from v=0 to v

Using the work-energy principle, the formula for the work needed to accelerate the object from v=0 to v is given by the following:

$\begin{array}{l}{\mathrm{W}}_1={\mathrm{KE}}_{1\mathrm{f}}-{\mathrm{KE}}_{1\mathrm{i}}\\ {\mathrm{W}}_1=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{1\mathrm{f}}}^2-\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{1\mathrm{i}}}^2\end{array}$

Here, m is the mass of the object, KE is the kinetic energy of the object, v is the velocity of the object, and subscripts i and f indicate the initial and final positions of the object.

Put ${\mathrm{v}}_{1\mathrm{i}}=0$and ${\mathrm{v}}_{1\mathrm{f}}=\mathrm{v}$in the formula.

$\begin{array}{l}{\mathrm{W}}_1=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}\mathrm{m}{\left(0\right)}^2\\ {\mathrm{W}}_1=\frac{1}{2}{\mathrm{mv}}^2...\left(1\right)\end{array}$

The work needed to accelerate the object from v to 2v

Similarly, the formula for the work required to accelerate the object from speed v to 2v is given by the following:

$\begin{array}{c}{W}_2=K{E}_{2f}-K{E}_{2i}\\ W_2=\frac{1}{2}m{v_{2f}}^2-\frac{1}{2}m{v_{2i}}^2\end{array}$

Put ${\mathrm{v}}_{2\mathrm{i}}=\mathrm{v}$and ${\mathrm{v}}_{2\mathrm{f}}=\mathrm{v}$in the formula.

$\begin{array}{c}{W}_2=\frac{1}{2}m{\left(2v\right)}^2-\frac{1}{2}m{v}^2\\ =\frac{1}{2}m\left(4v^2-v^2\right)\\ =3\times \frac{1}{2}m{v}^2\end{array}$

From equation (1),

${\mathrm{W}}_2=3\times {\mathrm{W}}_1$

Hence, the work required to accelerate the object from speed v to 2v is three times the work required to accelerate the object from v=0 to v.