Irodov 1.130: Half the Climb, Three Quarters the Work

Problem Irodov 1.130

A body of mass $m$ is pulled straight up from the Earth’s surface by an applied force $\vect F$ , directed upward, whose magnitude varies with the height already climbed, $y$ , as $F(y) = 2(1-ay)mg$ , where $a$ is a given positive constant and $g$ is the free-fall acceleration. The body starts from rest at $y=0$ and rises until its speed returns to zero at some height $H$ . Find the work $A$ done by $\vect F$ and the increment $\Delta U$ of the body’s gravitational potential energy, both taken over the first half of this total ascent, from $y=0$ to $y=H/2$ .

Free body diagram of the body at a general height y during the climb: the applied force F (gold) points upward and the weight m g (royal) points downward, both concurrent at the body since it is a point mass. — Free body diagram of the body at a general height $y$ during the climb: the applied force $\vect F$ (gold) points upward and the weight $m\vect g$ (royal) points downward, both concurrent at the body since it is a point mass.

Given

$m$
$\vect F(y) = 2(1-ay)mg$ , upward
$a>0$
$v=0$ at $y=0$

Find

the work

A

done by

\vect F

and the increment

\Delta U

of gravitational potential energy, each over the first half of the ascent

Solution

Step 1Net upward force as a function of height

Newton’s second law along the vertical combines the applied force with the weight to give the net force driving the ascent at any height $y$ .

\begin{align*}F_{\text{net}}(y) &= F(y) - mg \\ &= 2(1-ay)mg - mg \\ &= mg(1-2ay)\end{align*}

F_{\text{net}}(y) = mg(1-2ay)

(1)

Step 2Total height of the ascent

The work-energy theorem integrates the net force found in (1) from $y=0$ (at rest) to a general height $y$ to give the speed there; setting that speed to zero at the top of the climb, $y=H$ , isolates the total ascent height.

\begin{align*}\frac12 mv(y)^2 &= \int_0^y F_{\text{net}}(y')\,dy' \\ &= \int_0^y mg(1-2ay')\,dy' \\ &= mg\Bigl[y'-ay'^2\Bigr]_0^y \\ &= mg\left(y-ay^2\right)\end{align*}

Setting the left side to zero at $y=H>0$ ,

0 = mg\left(H-aH^2\right) = mgH(1-aH).

H = \dfrac{1}{a}

(2)

Step 3Work done by

\vect F

over the first half of the ascent

The work done by the applied force is the integral of $F(y)$ over the displacement, taken from launch to half the total ascent height found in (2).

\begin{align*}A &= \int_0^{H/2} F(y)\,dy \\ &= \int_0^{1/(2a)} 2(1-ay)mg\,dy \\ &= 2mg\left[y-\frac{ay^2}{2}\right]_0^{1/(2a)} \\ &= 2mg\left(\frac{1}{2a}-\frac{1}{8a}\right)\end{align*}

A = \dfrac{3mg}{4a}

(3)

Step 4Increment of gravitational potential energy

Gravitational potential energy depends only on height, so its increment over the same rise is $mg$ times the height climbed, regardless of how $F$ varied on the way up.

\Delta U = mg\cdot\frac{H}{2}

Substituting the ascent height $H=1/a$ from (2) gives the increment over the first half of the climb.

\Delta U = mg\cdot\dfrac{1}{2a} = \dfrac{mg}{2a}

(4)

Check

By the work-energy theorem the kinetic energy at $y=H/2$ equals the total work done there, $A-\Delta U = 3mg/4a - mg/2a = mg/4a$ . Independently, Step 2’s $v(y)^2=2g(y-ay^2)$ at $y=1/(2a)$ gives $v^2 = 2g(1/(2a)-1/(4a))=g/(2a)$ , so $mv^2/2 = mg/4a$ : the two agree.

Remark

The net force $F_{\text{net}}(y)=mg(1-2ay)$ vanishes exactly at $y=1/(2a)=H/2$ , so the body’s speed peaks precisely at the midpoint of its climb, not merely somewhere within the first half. That split at $H/2$ is not tied to the coefficient $2a$ : for any net force that falls off linearly with height and is positive at $y=0$ , the work done up to a general height is the integral of a linear function, hence a quadratic in that height whose vertex sits exactly where the force changes sign, and a quadratic that starts from zero has its second zero, which is the stopping height $H$ , at exactly twice its vertex. So the height where the net force itself vanishes, where the speed peaks, is always exactly $H/2$ for such a force, whatever its coefficients happen to be; this problem’s $2a$ is one instance of that general fact, not an exception to it.