Newton's Second Law — Question 1

Section 5-5 · Physics pages 692–693

Question 1 A net force determines acceleration, not velocity Use ∑⃗F = m⃗a for constant mass in an inertial frame. A 2 kg body has forces (8, 0) N and (−2, 4) N. Its initial velocity is (−1, 2) m/s. The forces remain constant for 3 s. 2 kg⃗ F1 = (8, 0) N⃗ F2 = (−2, 4) N⃗ v0 = (−1, 2) m/s Schematic; labels specify the data. Arrow lengths are not a common scale. (a) Find net force and acceleration. (b) Find velocity at 3 s. (c) Find displacement and explain why it is not generally parallel to the force.
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Question 1 - Solution A net force determines acceleration, not velocity 2 kg⃗ F1 = (8, 0) N⃗ F2 = (−2, 4) N⃗ v0 = (−1, 2) m/s (a) Sum before dividing.⃗ F = (8, 0) + (−2, 4) = (6, 4) N,⃗ a =⃗F/2 = (3, 2) m/s2⃗ Fnet = (6, 4) N, so⃗a = (3, 2) m/s2. (b) Add the velocity change.⃗ v(3) =⃗v0 +⃗a(3) = (−1, 2) + (9, 6) = (8, 8) m/s⃗ v(3) = (−1, 2) + (3, 2)3 = (8, 8) m/s. (c) Include the initial-motion term. ∆⃗r =⃗v0(3) + 1 2⃗a(3)2 = (−3, 6) + (13.5, 9) = (10.5, 15) m ∆⃗r =⃗v0t + 1 2⃗at2 = (−3, 6) + (13.5, 9) = (10.5, 15) m. The acceleration is parallel to net force; velocity and displacement also depend on initial data and need not share that direction.

Question 1

A net force determines acceleration, not velocity

Use ∑F→=ma→\sum\vec F=m\vec a for constant mass in an inertial frame. A 2 kg body has forces (8,0)(8,0) N and (−2,4)(-2,4) N. Its initial velocity is (−1,2)(-1,2) m/s. The forces remain constant for 3 s.

Diagram shown in the original SVG worksheet.

  1. Find net force and acceleration.

  2. Find velocity at 3 s.

  3. Find displacement and explain why it is not generally parallel to the force.

Question 1 - Solution

A net force determines acceleration, not velocity

Diagram shown in the original SVG worksheet.

(a) Sum before dividing.

F→=(8,0)+(−2,4)=(6,4)N,a→=F→/2=(3,2)m/s2\vec F=(8,0)+(-2,4)=(6,4)\,\mathrm N,\quad\boxed{\vec a=\vec F/2=(3,2)\,\mathrm{m/s^2}} F→net=(6,4)\vec F_{\mathrm{net}}=(6,4) N, so a→=(3,2)\vec a=(3,2) m/s2^2.

(b) Add the velocity change.

v→(3)=v→0+a→(3)=(−1,2)+(9,6)=(8,8)m/s\vec v(3)=\vec v_0+\vec a(3)=(-1,2)+(9,6)=(8,8)\,\mathrm{m/s} v→(3)=(−1,2)+(3,2)3=(8,8)\vec v(3)=(-1,2)+(3,2)3=(8,8) m/s.

(c) Include the initial-motion term.

Δr→=v→0(3)+12a→(3)2=(−3,6)+(13.5,9)=(10.5,15)m\Delta\vec r=\vec v_0(3)+\tfrac12\vec a(3)^2=(-3,6)+(13.5,9)=(10.5,15)\,\mathrm m Δr→=v→0t+12a→t2=(−3,6)+(13.5,9)=(10.5,15)\Delta\vec r=\vec v_0t+\tfrac12\vec at^2=(-3,6)+(13.5,9)=(10.5,15) m. The acceleration is parallel to net force; velocity and displacement also depend on initial data and need not share that direction.