Time-Varying Forces
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
When the force changes with time, F = ma still holds at every instant, so write m dv/dt = F(t) and integrate. The integral of force over time is the impulse, and it equals the change in momentum: J = ∫F dt = Δp.
1. Integrating Newton’s second law
2. Reading the motion from F(t)
While the net force is positive (along the motion), the object speeds up. Its velocity peaks when F crosses zero and turns negative.
If x(t) is given instead, differentiate twice to find a, then F = ma.
A changing force means a changing acceleration, so the constant-acceleration equations don’t apply. Integrate instead.
3. Example, then try it
A net force F = 4t² (N) acts on a 2.0 kg object starting from rest at x = 0. Find its velocity and position at t = 3.0 s.
a = F/m = 2t², so v = ∫2t² dt = (2/3)t³, and v(3) = 18 m/s.
x = ∫(2/3)t³ dt = t⁴/6, so x(3) = 81 ÷ 6 = 13.5 m.
The force in the graph below acts on a 3.0 kg object already moving at 2.0 m/s in the force’s direction. Find its velocity at t = 3 s and at t = 5 s.
Impulse to 3 s: triangle area ½(3)(12) = 18 N·s, so Δv = 18 ÷ 3.0 = 6.0 m/s.
v at 3 s: 2.0 + 6.0 = 8.0 m/s.
Impulse 3–5 s: (2)(12) = 24 N·s, so Δv = 8.0 m/s more.
v at 5 s: 8.0 + 8.0 = 16 m/s. (Total: 42 N·s ÷ 3.0 kg = 14 m/s gained.)
Answer: 8.0 m/s at 3 s and 16 m/s at 5 s.
4. Common mistakes
The net force on a moving object drops to zero and stays zero. What does the object do afterward?
With zero net force, dv/dt = 0, so the velocity stays at whatever it was when the force reached zero.
Ready to practice?
5 questions on Time-Varying Forces, each with a hint and an explanation for every wrong answer.