A body of mass m = 50 kg moves at v_0 = 4 m/s . A constant…

A body of mass $m = 50\ \mathrm{kg}$ moves at $v_0 = 4\ \mathrm{m/s}$. A constant force $F = 30\ \mathrm{N}$ acts on it for $t = 2\ \mathrm{s}$, opposite to its motion. (a) By how much does its speed decrease? (b) How large would the force have to be to stop the body in the same time?

Explanation of Method

Slowing and stopping a moving body.

Step 1: (a) Decrease in speed

By the impulse–momentum theorem: \[F t = m\,\Delta v \;\Longrightarrow\; \Delta v = \frac{Ft}{m} = \frac{30 \cdot 2}{50}.\]

\[\Delta v = 1.2\ \mathrm{m/s} \ \ (\text{from } 4 \text{ to } 2.8\ \mathrm{m/s})\]

Step 2: (b) Stopping force

\[ F t = m v_0 \;\Longrightarrow\; F = \frac{m v_0}{t} = \frac{50 \cdot 4}{2}. \]

\[F = 100\ \mathrm{N} \ \text{(against the motion)}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $2.8$ $4$ $F = 30\ \mathrm{N}: \ \Delta v = 1.2\ \mathrm{m/s}$ $F = 100\ \mathrm{N}: \text{ stops}$

Speed against time. The braking force sets the slope $F/m$ of the line: 30 N removes 1.2 m/s in 2 s, while stopping from 4 m/s in 2 s needs 100 N.

Result

\[\boxed{\,\Delta v = 1.2\ \mathrm{m/s} \ \ (\text{from } 4 \text{ to } 2.8\ \mathrm{m/s})\,}\]

\[\boxed{\,F = 100\ \mathrm{N} \ \text{(against the motion)}\,}\]

Remark

Check. The original results are correct.