Problem 248
Easy Translational Dynamics

A constant force F = 1 N gives a body an acceleration a = 10…

A constant force $F = 1\ \mathrm{N}$ gives a body an acceleration $a = 10\ \mathrm{cm/s^2}$. If the body starts from rest, what is its kinetic energy after $t = 5\ \mathrm{s}$?

Explanation of Method

Kinetic energy after 5 s of a constant force.

Step 1: Mass and speed

\[ m = \frac{F}{a} = \frac{1}{0.1} = 10\ \mathrm{kg}, \]

\[ v = at = 0.5\ \mathrm{m/s}. \]

Step 2: Kinetic energy

\[E_k = \tfrac12 m v^2 = \tfrac12 \cdot 10 \cdot 0.25 = 1.25\ \mathrm{J}\]

Check

Work: $ s = \tfrac12 at^2 = 1.25\ \mathrm{m} $, $ A = Fs = 1.25\ \mathrm{J}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $4$ $5$ $0.5$ $E_k = \tfrac12 m v^2 = 1.25\ \mathrm{J}$ $s = 1.25\ \mathrm{m}$
Speed against time for the 10 kg body. The area under the line is the distance, 1.25 m; the work $F s = 1\ \mathrm{N} \times 1.25\ \mathrm{m}$ equals the kinetic energy gained.

Result

\[\boxed{\,E_k = \tfrac12 m v^2 = \tfrac12 \cdot 10 \cdot 0.25 = 1.25\ \mathrm{J}\,}\]