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Physics / Chapter 7: Kinetic Energy and Work / Section 7-8

Power — Question 8

Section 7-8 · Physics pages 1006–1007

Question 8 Regenerative braking separates removed kinetic energy from recovered energy A 1000 kg vehicle slows from 20 to 10 m/s in 5 s. Braking is its only agent of net work; 60 percent of the kinetic energy removed is recovered electrically. Ignore changes in height. vehicle braking recovered unrecovered The removed mechanical energy splits into recovered electrical energy and other output. (a) Find braking work and average mechanical braking power. (b) Find recovered electrical energy and average recovery power. (c) Can the instantaneous braking power be inferred from these endpoint data?
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Question 8 - Solution Regenerative braking separates removed kinetic energy from recovered energy vehicle braking recovered unrecovered (a) Keep mechanical work negative. The governing relation is Wb = ∆K, Erecovered = −ηWb Wb = 1 2(1000)(102 −202) = −150000 J. Average power delivered to the vehicle is −30000 W. (b) Apply the recovery fraction to the removed amount. Recovered energy is .60(150000) = 90000 J. Average electrical recovery is 18000 W, with the remaining 60000 J unrecovered. (c) Distinguish an average from a history. No. Many speed and force histories have the same endpoints and duration. They share the average mechanical power but need not share instantaneous power or peak load.

Question 8

Regenerative braking separates removed kinetic energy from recovered energy

A 1000 kg vehicle slows from 20 to 10 m/s in 5 s. Braking is its only agent of net work; 60 percent of the kinetic energy removed is recovered electrically. Ignore changes in height.

Diagram shown in the original SVG worksheet.

  1. Find braking work and average mechanical braking power.

  2. Find recovered electrical energy and average recovery power.

  3. Can the instantaneous braking power be inferred from these endpoint data?

Question 8 - Solution

Regenerative braking separates removed kinetic energy from recovered energy

Diagram shown in the original SVG worksheet.

(a) Keep mechanical work negative.

The governing relation is Wb=ΔK,Erecovered=−ηWb\boxed{W_b=\Delta K,\qquad E_{\mathrm{recovered}}=-\eta W_b}

Wb=12(1000)(102−202)=−150000J.W_b=\tfrac12(1000)(10^2-20^2)=-150000\ \mathrm J. Average power delivered to the vehicle is −30000W-30000\ \mathrm W.

(b) Apply the recovery fraction to the removed amount.

Recovered energy is .60(150000)=90000J..60(150000)=90000\ \mathrm J. Average electrical recovery is 18000 W, with the remaining 60000 J unrecovered.

(c) Distinguish an average from a history.

No. Many speed and force histories have the same endpoints and duration. They share the average mechanical power but need not share instantaneous power or peak load.

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