Work Done on a System by an External Force — Question 3

Section 8-6 · Physics pages 1116–1117

Question 3 Uniform ascent with internal frictional heating Assume no heat transfer across the stated system boundary during the process unless otherwise specified. A 3 kg crate is pulled at constant speed up a fixed incline for 4 m, rising vertically 2 m. Friction is 5 N. Choose crate, ramp and Earth as the system; take g=10 m/s2. 2 m 4 m along ramp crate The incline length and vertical rise are different distances in the balance. (a) Find changes in kinetic, gravitational and thermal energies. (b) Find external pulling work. (c) Find the constant pulling force if it acts parallel to the incline.
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Question 3 - Solution Uniform ascent with internal frictional heating 2 m 4 m along ramp crate (a) Identify all changing stores. Constant speed eliminates the kinetic change, but not the other stores: ∆K = 0 , ∆Ug = 3(10)(2) = 60 J , ∆Eth = 5(4) = 20 J . (b) Sum the energy requirements. The external input equals the total storage increase: Wpull = ∆K + ∆Ug + ∆Eth = 0 + 60 + 20 = 80 J . (c) Relate the external work to its own displacement. The force acts along its 4 m displacement: Fpull = Wpull/s = 80/4 = 20 N . Force balance confirms mg sin θ + f = 30(2/4) + 5 = 20 N.

Question 3

Uniform ascent with internal frictional heating

Assume no heat transfer across the stated system boundary during the process unless otherwise specified. A 3 kg crate is pulled at constant speed up a fixed incline for 4 m, rising vertically 2 m. Friction is 5 N. Choose crate, ramp and Earth as the system; take g=10 m/s2^2.

Diagram shown in the original SVG worksheet.

  1. Find changes in kinetic, gravitational and thermal energies.

  2. Find external pulling work.

  3. Find the constant pulling force if it acts parallel to the incline.

Question 3 - Solution

Uniform ascent with internal frictional heating

Diagram shown in the original SVG worksheet.

(a) Identify all changing stores.

Constant speed eliminates the kinetic change, but not the other stores: ΔK=0,ΔUg=3(10)(2)=60J,\boxed{\Delta K=0},\qquad\boxed{\Delta U_g=3(10)(2)=60\ \mathrm J}, ΔEth=5(4)=20J.\boxed{\Delta E_{\mathrm{th}}=5(4)=20\ \mathrm J}.

(b) Sum the energy requirements.

The external input equals the total storage increase: Wpull=ΔK+ΔUg+ΔEth=0+60+20=80J.\boxed{W_{\mathrm{pull}}=\Delta K+\Delta U_g+\Delta E_{\mathrm{th}}=0+60+20=80\ \mathrm J}.

(c) Relate the external work to its own displacement.

The force acts along its 4 m displacement: Fpull=Wpull/s=80/4=20N.\boxed{F_{\mathrm{pull}}=W_{\mathrm{pull}}/s=80/4=20\ \mathrm N}. Force balance confirms mgsin⁡θ+f=30(2/4)+5=20mg\sin\theta+f=30(2/4)+5=20 N.