Stem-Master
Physics / Chapter 6: Force and Motion - II / Section 6-3

The Drag Force and Terminal Speed — Question 8

Section 6-3 · Physics pages 826–827

Question 8 Buoyancy can reverse the direction of terminal motion Take g = 10 m/s2 and neglect buoyancy unless explicitly specified. A small body has mass 0.30 kg and displaces a fixed fluid volume giving a constant upward buoyant force B=4 N. It is fully submerged in still fluid. With upward velocity v positive, drag is Fd = −bv, where b = 0.50 kg/s. B mg positive Buoyancy is supplied as a force; the drag direction must be inferred from the signed velocity. (a) Find terminal velocity and direction. (b) Find acceleration when the body is moving downward at 2 m/s. (c) If B were changed to 2 N with all else unchanged, find the new terminal velocity and explain the sign change.
Show solutionHide solution
Question 8 - Solution Buoyancy can reverse the direction of terminal motion B mg positive (a) Balance signed forces. With upward positive, the complete equation is m˙v = B −mg −bv. Setting ˙v = 0 gives vt = B −mg b = 4 −.30(10) .50 = +2 m/s upward. (b) Use negative v for downward motion. Downward motion at 2 m/s means v = −2 m/s. Thus Fd = −bv = −.50(−2) = +1 N, a = 4 −3 + 1 .30 = 20 3 = 6.667 m/s2 upward . (c) Change the effective driving force. For the changed buoyancy, vt = 2 −3 .50 = −2 m/s downward. The sign follows B −mg. A speed-only expression would lose the distinction between rising and sinking.

Question 8

Buoyancy can reverse the direction of terminal motion

Take g=10g=10 m/s2^2 and neglect buoyancy unless explicitly specified. A small body has mass 0.30 kg and displaces a fixed fluid volume giving a constant upward buoyant force B=4 N. It is fully submerged in still fluid. With upward velocity v positive, drag is Fd=−bvF_d=-bv, where b=0.50b=0.50 kg/s.

Diagram shown in the original SVG worksheet.

  1. Find terminal velocity and direction.

  2. Find acceleration when the body is moving downward at 2 m/s.

  3. If B were changed to 2 N with all else unchanged, find the new terminal velocity and explain the sign change.

Question 8 - Solution

Buoyancy can reverse the direction of terminal motion

Diagram shown in the original SVG worksheet.

(a) Balance signed forces.

With upward positive, the complete equation is mv̇=B−mg−bv.m\dot v=B-mg-bv. Setting v̇=0\dot v=0 gives vt=B−mgb=4−.30(10).50=+2m/supward.\boxed{v_t=\frac{B-mg}{b}=\frac{4-.30(10)}{.50}=+2\ \mathrm{m/s}}\quad\text{upward}.

(b) Use negative v for downward motion.

Downward motion at 2 m/s means v=−2v=-2 m/s. Thus Fd=−bv=−.50(−2)=+1N,F_d=-bv=-.50(-2)=+1\ \mathrm N, a=4−3+1.30=203=6.667m/s2 upward.\boxed{a=\frac{4-3+1}{.30}=\frac{20}{3}=6.667\ \mathrm{m/s^2}\text{ upward}}.

(c) Change the effective driving force.

For the changed buoyancy, vt=2−3.50=−2m/sdownward.\boxed{v_t=\frac{2-3}{.50}=-2\ \mathrm{m/s}}\quad\text{downward}. The sign follows B−mgB-mg. A speed-only expression would lose the distinction between rising and sinking.

← Section questionsChapter contents