Newton's Second Law — Question 6

Question 6

A time-dependent force must be integrated

Use $\sum\vec F=m\vec a$ for constant mass in an inertial frame. A 2 kg body starts at rest at x=0. For $t\ge0$, the net force is $F_x=(4\ \mathrm{N/s})t$.

1. Find acceleration and velocity as functions of time.

2. Find position at 3 s.

3. Why is using the final acceleration over the entire interval wrong?

Question 6 - Solution

A time-dependent force must be integrated

(a) Integrate the actual rate.

$$a(t)=\frac{4t}{2}=2t,\quad v(t)=\int_0^t2u\,du=t^2\quad\text{(SI)}$$ $a_x=(2\ \mathrm{m/s^3})t$. Integrating from zero gives $v_x=(1\ \mathrm{m/s^3})t^2$.

(b) Integrate once more.

$$x(t)=\int_0^tu^2\,du=\frac{t^3}{3},\quad\boxed{x(3)=9\,\mathrm m}$$ $x=(1/3\ \mathrm{m/s^3})t^3$, so $x(3\ \mathrm s)=9$ m and $v(3\ \mathrm s)=9$ m/s.

(c) Respect the time history.

$$a(3)=6,\quad v_{\mathrm{wrong}}=6(3)=18\,\mathrm{m/s},\quad x_{\mathrm{wrong}}=\tfrac12(6)(3)^2=27\,\mathrm m$$ The final acceleration is 6 m/s$^2$. Treating it as constant for 3 s would predict 18 m/s and 27 m, neither correct. The body spent earlier times under a smaller force.