Newton's First Law — Question 1

Section 5-2 · Physics pages 632–633

Question 1 Coasting does not require a forward net force A puck moves at 3 m/s east on an ideal horizontal surface. At t=0 the brief push ends. All remaining forces balance. Treat the ground as inertial. x Equal-time positions with equal spacing represent constant velocity. (a) Find its velocity and displacement over the next 4 s. (b) Does the end of the push make it gradually stop in this model? (c) If a real puck slows, what assumption needs revision?
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Question 1 - Solution Coasting does not require a forward net force x (a) Apply zero net force to velocity. ∑⃗ F =⃗0 ⇒⃗a =⃗0, vx = 3 m/s, ∆x = 3(4) = 12 m The first law gives constant velocity, 3 m/s east. Displacement in 4 s is 12 m east. (b) Do not confuse force with velocity. vx(t) = vx(0) + ∫t 0 0 du = vx(0) No forward net force is needed to maintain the existing velocity. The stipulated model contains no unbalanced interaction that could change it. (c) Reopen the interaction inventory. dvx dt < 0 (vx > 0) ⇒Fnet,x = mdvx dt < 0 Slowing would require nonzero acceleration and thus failure of the balanced-force assumption in an inertial frame. Surface resistance or air interaction could explain it. That physical evidence revises the idealization, not the first law itself.

Question 1

Coasting does not require a forward net force

A puck moves at 3 m/s east on an ideal horizontal surface. At t=0 the brief push ends. All remaining forces balance. Treat the ground as inertial.

Diagram shown in the original SVG worksheet.

  1. Find its velocity and displacement over the next 4 s.

  2. Does the end of the push make it gradually stop in this model?

  3. If a real puck slows, what assumption needs revision?

Question 1 - Solution

Coasting does not require a forward net force

Diagram shown in the original SVG worksheet.

(a) Apply zero net force to velocity.

∑F→=0→⇒a→=0→,vx=3m/s,Δx=3(4)=12m\sum\vec F=\vec0\ \Rightarrow\ \vec a=\vec0,\quad \boxed{v_x=3\,\mathrm{m/s},\quad\Delta x=3(4)=12\,\mathrm m} The first law gives constant velocity, 3 m/s east. Displacement in 4 s is 12 m east.

(b) Do not confuse force with velocity.

vx(t)=vx(0)+∫0t0du=vx(0)v_x(t)=v_x(0)+\int_0^t0\,du=v_x(0) No forward net force is needed to maintain the existing velocity. The stipulated model contains no unbalanced interaction that could change it.

(c) Reopen the interaction inventory.

dvxdt<0(vx>0)⇒Fnet,x=mdvxdt<0\frac{dv_x}{dt}<0\ (v_x>0)\ \Rightarrow\ F_{{\mathrm{net}},x}=m\frac{dv_x}{dt}<0 Slowing would require nonzero acceleration and thus failure of the balanced-force assumption in an inertial frame. Surface resistance or air interaction could explain it. That physical evidence revises the idealization, not the first law itself.