Work Done on a System by an External Force — Question 4

Section 8-6 · Physics pages 1118–1119

Question 4 Controlled descent with energy extracted by an agent Assume no heat transfer across the stated system boundary during the process unless otherwise specified. A 2 kg load descends 3 m at constant speed against a 4 N drag force. Choose load, Earth and the drag medium as the system. A separate controlling cable is external. Use g=10 m/s2. load T drag mg Cable tension and drag point upward during the controlled downward motion. (a) Find gravitational and thermal energy changes. (b) Find cable work on the system. (c) Find cable tension and explain the direction of energy transfer.
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Question 4 - Solution Controlled descent with energy extracted by an agent load T drag mg (a) Record storage changes with signs. Record the signed changes for the descending load: ∆Ug = mg∆y = 2(10)(−3) = −60 J , ∆Eth = 4(3) = 12 J . (b) Use work equal to net system-energy change. Kinetic energy remains unchanged: WT = ∆K + ∆Ug + ∆Eth = 0 −60 + 12 = −48 J . (c) Cross-check forces and work. Upward force balance gives T + 4 −mg = 0 ⇒T = 16 N , WT = T(−3) = −48 J. The controller receives 48 J and the medium retains 12 J thermally, accounting for the full 60 J potential decrease.

Question 4

Controlled descent with energy extracted by an agent

Assume no heat transfer across the stated system boundary during the process unless otherwise specified. A 2 kg load descends 3 m at constant speed against a 4 N drag force. Choose load, Earth and the drag medium as the system. A separate controlling cable is external. Use g=10 m/s2^2.

Diagram shown in the original SVG worksheet.

  1. Find gravitational and thermal energy changes.

  2. Find cable work on the system.

  3. Find cable tension and explain the direction of energy transfer.

Question 4 - Solution

Controlled descent with energy extracted by an agent

Diagram shown in the original SVG worksheet.

(a) Record storage changes with signs.

Record the signed changes for the descending load: ΔUg=mgΔy=2(10)(−3)=−60J,ΔEth=4(3)=12J.\boxed{\Delta U_g=mg\Delta y=2(10)(-3)=-60\ \mathrm J},\qquad\boxed{\Delta E_{\mathrm{th}}=4(3)=12\ \mathrm J}.

(b) Use work equal to net system-energy change.

Kinetic energy remains unchanged: WT=ΔK+ΔUg+ΔEth=0−60+12=−48J.\boxed{W_T=\Delta K+\Delta U_g+\Delta E_{\mathrm{th}}=0-60+12=-48\ \mathrm J}.

(c) Cross-check forces and work.

Upward force balance gives T+4−mg=0⇒T=16N,WT=T(−3)=−48J.T+4-mg=0\Rightarrow\boxed{T=16\ \mathrm N},\qquad W_T=T(-3)=-48\ \mathrm J. The controller receives 48 J and the medium retains 12 J thermally, accounting for the full 60 J potential decrease.