Under the action of a constant force F , a body of mass m =…

Under the action of a constant force $F$, a body of mass $m = 50\ \mathrm{g}$, starting from rest, travels $s = 30\ \mathrm{m}$ in $t = 2\ \mathrm{s}$. How large is the force?

Explanation of Method

Force from distance and time.

Step 1: Acceleration

\[ s = \tfrac12 a t^2 \;\Longrightarrow\; a = \frac{2s}{t^2} = \frac{60}{4} = 15\ \mathrm{m/s^2}. \]

Step 2: Force

\[ F = ma = 0.05 \cdot 15. \]

\[F = 0.75\ \mathrm{N}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $15$ $30$ $s = 30\ \mathrm{m}$ $a = 15\ \mathrm{m/s^2}$

Speed against time for the uniformly accelerated body: the area under the line is the 30 m covered in 2 s.

Result

\[\boxed{\,F = 0.75\ \mathrm{N}\,}\]

Remark

Check. The original result is correct (assuming $F$ is the net force, e.g. on a frictionless horizontal surface). The body's final speed is $at = 30\ \mathrm{m/s}$.