Newton's Second Law — Question 3

Section 5-5 · Physics pages 696–697

Question 3 Recover an unknown force from a measured trajectory Use ∑⃗F = m⃗a for constant mass in an inertial frame. A 4 kg particle follows⃗r = (t2, 3t −t2/2) m for numerical t in seconds. One force is (3, 5) N; one other force acts. x (m) y (m) 4 kg⃗ F1 = (3, 5) N Parametric path for 0 to 3 s; the identified force is shown schematically. (a) Find the required net force. (b) Find the other force. (c) Check whether the particle’s velocity must be parallel to that net force.
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Question 3 - Solution Recover an unknown force from a measured trajectory x (m) y (m) 4 kg⃗ F1 = (3, 5) N (a) Differentiate twice.⃗ v = (2t, 3 −t),⃗ a = (2, −1),⃗ Fnet = 4(2, −1) = (8, −4) N⃗ v = (2t, 3 −t) m/s and⃗a = (2, −1) m/s2. Thus⃗Fnet = (8, −4) N. (b) Subtract the known interaction.⃗ F2 =⃗Fnet −⃗F1 = (8, −4) −(3, 5) = (5, −9) N⃗ F2 = (8, −4) −(3, 5) = (5, −9) N. (c) Test a concrete instant. vxFy −vyFx = (2t)(−4) −(3 −t)(8) = −24̸ = 0 At t=0, velocity is (0, 3) m/s, not parallel to (8, −4) N. Newton’s second law aligns acceleration with the resultant, not velocity.

Question 3

Recover an unknown force from a measured trajectory

Use ∑F→=ma→\sum\vec F=m\vec a for constant mass in an inertial frame. A 4 kg particle follows r→=(t2,3t−t2/2)\vec r=(t^2,3t-t^2/2) m for numerical t in seconds. One force is (3,5)(3,5) N; one other force acts.

Diagram shown in the original SVG worksheet.

  1. Find the required net force.

  2. Find the other force.

  3. Check whether the particle’s velocity must be parallel to that net force.

Question 3 - Solution

Recover an unknown force from a measured trajectory

Diagram shown in the original SVG worksheet.

(a) Differentiate twice.

v→=(2t,3−t),a→=(2,−1),F→net=4(2,−1)=(8,−4)N\vec v=(2t,3-t),\quad\vec a=(2,-1),\quad\boxed{\vec F_{\mathrm{net}}=4(2,-1)=(8,-4)\,\mathrm N} v→=(2t,3−t)\vec v=(2t,3-t) m/s and a→=(2,−1)\vec a=(2,-1) m/s2^2. Thus F→net=(8,−4)\vec F_{\mathrm{net}}=(8,-4) N.

(b) Subtract the known interaction.

F→2=F→net−F→1=(8,−4)−(3,5)=(5,−9)N\vec F_2=\vec F_{\mathrm{net}}-\vec F_1=(8,-4)-(3,5)=(5,-9)\,\mathrm N F→2=(8,−4)−(3,5)=(5,−9)\vec F_2=(8,-4)-(3,5)=(5,-9) N.

(c) Test a concrete instant.

vxFy−vyFx=(2t)(−4)−(3−t)(8)=−24≠0v_xF_y-v_yF_x=(2t)(-4)-(3-t)(8)=-24\ne0 At t=0, velocity is (0,3)(0,3) m/s, not parallel to (8,−4)(8,-4) N. Newton’s second law aligns acceleration with the resultant, not velocity.