A body of mass m = 2 kg is lifted vertically by a constant…

A body of mass $m = 2\ \mathrm{kg}$ is lifted vertically by a constant force $F$. While raising the body by $h = 1\ \mathrm{m}$, the force does work $A = 78.5\ \mathrm{J}$. Find the acceleration.

Explanation of Method

Acceleration from the work of a lifting force.

Step 1: The force

\[ A = Fh \;\Longrightarrow\; F = \frac{78.5}{1} = 78.5\ \mathrm{N}. \]

Step 2: Acceleration

\[ F - mg = ma \;\Longrightarrow\; a = \frac{F}{m} - g = 39.25 - 9.81. \]

\[a \approx 29.4\ \mathrm{m/s^2}\]

$F = \frac{A}{h} = 78.5\ \mathrm{N}$ $mg = 19.6\ \mathrm{N}$ $h = 1\ \mathrm{m}$ $a = \frac{F - mg}{m} \approx 29.4\ \mathrm{m/s^2}$

The work of the force over 1 m gives the force itself, 78.5 N. Most of it goes into accelerating the 2 kg body; only 19.6 N balances its weight.

Result

\[\boxed{\,a \approx 29.4\ \mathrm{m/s^2}\,}\]

Remark

Check. The original result is correct (29.45 m/s² with $g = 9.8$). Of the 78.5 J, 19.6 J become potential energy and 58.9 J kinetic energy.