Newton's First Law — Question 2

Section 5-2 · Physics pages 634–635

Question 2 Balanced forces allow either rest or steady motion A box is subject to 12 N east and 12 N west horizontally; its vertical forces also balance. Consider one run initially at rest and another initially moving west at 2 m/s. 12 N 12 N The two horizontal forces act on the same box and cancel. (a) Find the horizontal net force. (b) Predict each run’s displacement after 5 s. (c) Can a force inventory with zero resultant determine the initial velocity?
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Question 2 - Solution Balanced forces allow either rest or steady motion 12 N 12 N (a) Add signed forces. Fnet,x = 12 −12 = 0 N , ax = 0 With east positive, Fnet,x = 12 −12 = 0 N. The full resultant is zero. (b) Preserve each initial velocity. ∆xA = 0(5) = 0, ∆xB = (−2)(5) = −10 m The resting run remains at its starting point. The moving run displaces 10 m west in 5 s. Both satisfy the same balanced-force condition. (c) Supply the missing initial state. ˙v = 0 ⇒v(t) = C, C = v(0) The first law specifies that velocity remains constant when net force vanishes. It does not assign the value of that constant. An initial motion measurement is required.

Question 2

Balanced forces allow either rest or steady motion

A box is subject to 12 N east and 12 N west horizontally; its vertical forces also balance. Consider one run initially at rest and another initially moving west at 2 m/s.

Diagram shown in the original SVG worksheet.

  1. Find the horizontal net force.

  2. Predict each run’s displacement after 5 s.

  3. Can a force inventory with zero resultant determine the initial velocity?

Question 2 - Solution

Balanced forces allow either rest or steady motion

Diagram shown in the original SVG worksheet.

(a) Add signed forces.

Fnet,x=12−12=0N,ax=0\boxed{F_{{\mathrm{net}},x}=12-12=0\,\mathrm N},\quad a_x=0 With east positive, Fnet,x=12−12=0F_{\mathrm{net},x}=12-12=0 N. The full resultant is zero.

(b) Preserve each initial velocity.

ΔxA=0(5)=0,ΔxB=(−2)(5)=−10m\Delta x_A=0(5)=0,\quad\Delta x_B=(-2)(5)=-10\,\mathrm m The resting run remains at its starting point. The moving run displaces 10 m west in 5 s. Both satisfy the same balanced-force condition.

(c) Supply the missing initial state.

v̇=0⇒v(t)=C,C=v(0)\dot v=0\ \Rightarrow\ v(t)=C,\quad C=v(0) The first law specifies that velocity remains constant when net force vanishes. It does not assign the value of that constant. An initial motion measurement is required.