Crate with a piecewise force

Problem

A $5\ \text{kg}$ crate starts from rest on a smooth floor. The applied force is $F = 10\ \text{N}$ between $t = 0$ and $t = 2\ \text{s}$, then $F = 0$ between $2$ and $4\ \text{s}$, and finally $F = -5\ \text{N}$ between $4$ and $6\ \text{s}$. Sketch $v(t)$ qualitatively and calculate $v(2)$, $v(4)$, $v(6)$.

Solution

The floor is smooth, so the only horizontal force is the applied force $F$. By Newton’s second law the acceleration in each interval is $a = F/m$, with $m = 5\ \text{kg}$. Let us analyse the three intervals, bearing in mind that the crate starts from rest, $v(0) = 0$.

Interval $0 \le t \le 2\ \text{s}$ ($F = 10\ \text{N}$): $a_1 = \frac{F}{m} = \frac{10}{5} = 2{,}0\ \text{m/s}^2$ The velocity grows linearly: $v(2) = v(0) + a_1 \cdot 2 = 0 + 2{,}0 \cdot 2 = 4\ \text{m/s}$

Interval $2 \le t \le 4\ \text{s}$ ($F = 0$): $a_2 = \frac{0}{5} = 0\ \text{m/s}^2$ With no force the crate proceeds by inertia at constant velocity: $v(4) = v(2) + a_2 \cdot 2 = 4 + 0 = 4\ \text{m/s}$

Interval $4 \le t \le 6\ \text{s}$ ($F = -5\ \text{N}$): $a_3 = \frac{-5}{5} = -1{,}0\ \text{m/s}^2$ The force is now opposite to the motion and the crate slows down: $v(6) = v(4) + a_3 \cdot 2 = 4 + (-1{,}0) \cdot 2 = 2\ \text{m/s}$

$\ev{v(2) = 4\ \text{m/s}, \quad v(4) = 4\ \text{m/s}, \quad v(6) = 2\ \text{m/s}}$

Qualitative graph of $v(t)$. The velocity–time diagram is made up of three straight segments: a rising ramp from $0$ to $4\ \text{m/s}$ in the first $2\ \text{s}$ (slope $+2$), a flat stretch at $4\ \text{m/s}$ between $2$ and $4\ \text{s}$ (zero slope), and finally a descent from $4$ to $2\ \text{m/s}$ between $4$ and $6\ \text{s}$ (slope $-1$). The slope of each segment is precisely the corresponding acceleration.

Links

Topics: Dynamics · Kinematics Concepts: Newton’s second law · Acceleration Skills: Graph reading