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Physics / Chapter 7: Kinetic Energy and Work / Section 7-4

Work and Kinetic Energy — Question 7

Section 7-4 · Physics pages 924–925

Question 7 Work and kinetic-energy change transform together A constant horizontal net force +2 N acts on a 2 kg cart for 2 s. Its initial ground velocity is +1 m/s. A second observer moves at constant +4 m/s. Use x′ = x −4t and v′ = v −4, with SI units. x initial: vi = 1 m/s final: vf = 3 m/s observer speed u Galilean transformation: v′ i = 1 −u and v′ f = 3 −u. (a) Find the ground displacement, work and kinetic-energy change. (b) Find the displacement and work in the moving frame. (c) Check the kinetic-energy change there and explain the sign reversal.
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Question 7 - Solution Work and kinetic-energy change transform together x initial: vi = 1 m/s final: vf = 3 m/s observer speed u (a) Use constant acceleration to determine the interval. The governing relation is ∆x′ = ∆x −u∆t, W ′ = F∆x′ a = 2/2 = 1 m/s2, so vf = 3 m/s, ∆x = 1(2) + 1 2(1)22 = 4 m. Work is 8 J, matching Kf −Ki = 9 −1 = 8 J. (b) Transform the interval displacement. ∆x′ = 4 −4(2) = −4 m. The force remains +2 N under a classical Galilean frame change, so W ′ = −8 J. (c) Transform both endpoint velocities. v′ i = −3, v′ f = −1 m/s, giving ∆K′ = 1 −9 = −8 J. The cart slows in this frame while speeding up in the ground frame; the theorem holds in both.

Question 7

Work and kinetic-energy change transform together

A constant horizontal net force +2 N acts on a 2 kg cart for 2 s. Its initial ground velocity is +1 m/s. A second observer moves at constant +4 m/s. Use x′=x−4tx^{\prime}=x-4t and v′=v−4v^{\prime}=v-4, with SI units.

Diagram shown in the original SVG worksheet.

  1. Find the ground displacement, work and kinetic-energy change.

  2. Find the displacement and work in the moving frame.

  3. Check the kinetic-energy change there and explain the sign reversal.

Question 7 - Solution

Work and kinetic-energy change transform together

Diagram shown in the original SVG worksheet.

(a) Use constant acceleration to determine the interval.

The governing relation is Δx′=Δx−uΔt,W′=FΔx′\boxed{\Delta x^\prime=\Delta x-u\Delta t,\qquad W^\prime=F\Delta x^\prime}

a=2/2=1m/s2,a=2/2=1\ \mathrm{m/s^2}, so vf=3m/s,Δx=1(2)+12(1)22=4m.v_f=3\ \mathrm{m/s},\qquad \Delta x=1(2)+\tfrac12(1)2^2=4\ \mathrm m. Work is 8 J, matching Kf−Ki=9−1=8J.K_f-K_i=9-1=8\ \mathrm J.

(b) Transform the interval displacement.

Δx′=4−4(2)=−4m.\Delta x^{\prime}=4-4(2)=-4\ \mathrm m. The force remains +2 N under a classical Galilean frame change, so W′=−8J.W^{\prime}=-8\ \mathrm J.

(c) Transform both endpoint velocities.

vi′=−3,vf′=−1m/s,v_i^{\prime}=-3,\qquad v_f^{\prime}=-1\ \mathrm{m/s}, giving ΔK′=1−9=−8J.\Delta K^{\prime}=1-9=-8\ \mathrm J. The cart slows in this frame while speeding up in the ground frame; the theorem holds in both.

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