3.2 Work — Worked Solutions
Work through Topic 3.2
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A constant force is perpendicular to a particle’s displacement. What work does it do?
- Fd
- Zero
- Negative Fd
Step 1. Use work W = Fd cos θ with θ = 90°.
Why this step? The problem gives a particle and a force perpendicular to its displacement, so this angle selects the force component along that displacement.
Step 2. Since cos 90° = 0, W = 0 J.
Why this step? There is no parallel force component through the displacement, so this interaction transfers no energy to the particle’s kinetic motion.
Step 3. The force may still change motion direction.
Why this step? A zero work result constrains kinetic-energy change, not acceleration; a perpendicular force can turn the velocity without changing speed.
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Question 2
What does area under an F_parallel-versus-displacement graph represent?
- Work
- Impulse
- Power
Step 1. For one constant-force graph segment, area is F_parallelΔx.
Why this step? The graph’s height is the force component doing work, and its width is the corresponding displacement interval.
Step 2. Sum the signed segment areas to find W.
Why this step? Successive displacement intervals contribute their own work, and below-axis forces must subtract rather than add positive area.
Step 3. The units are N·m = J, identifying work rather than impulse or power.
Why this step? Impulse would require a time-axis width, and power would divide energy by time; neither operation is present in this graph.
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Question 3
A box is dragged out and back along a rough floor. What can be true?
- Friction work is zero because displacement is zero
- Friction does negative work on both legs
- Friction is conservative
Step 1. On the outward leg, sliding friction points opposite displacement, so W_f,out = −f_kL_out for constant magnitude.
Why this step? The force-displacement angle is 180°, making its work negative even when the outward coordinate is positive.
Step 2. On the return leg, friction reverses direction with the sliding but still opposes displacement, so W_f,return = −f_kL_return.
Why this step? Changing the path direction also changes friction direction; it does not turn the friction work positive.
Step 3. Total friction work is −f_k(L_out + L_return), generally nonzero despite zero net displacement.
Why this step? This force is path-dependent; traveled length, not endpoint displacement, counts both negative energy transfers.
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Question 4
A 5 N horizontal force acts over 4 m while a 1 N opposing force acts throughout. For a 2 kg object starting at rest, find each work, net work, and final speed.
Step 1. Applied work is W_pull = 5·4 = +20 J.
Why this step? The 5 N force points along the 4 m displacement, so its parallel component contributes positive energy transfer.
Step 2. Opposing-force work is W_opp = −1·4 = −4 J.
Why this step? The opposing force has a 180° angle with displacement, giving a negative contribution rather than reducing the displacement itself.
Step 3. Net work is W_net = 20 − 4 = 16 J.
Why this step? The work-energy theorem uses the sum of all force work on the object; the applied-force work alone would overestimate the energy gain.
Step 4. From rest, K_i = 0 and ½(2)v_f² = 16, so v_f = 4 m/s.
Why this step? The object model converts net work into translational kinetic-energy change; the positive root is the requested speed magnitude.
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Question 5
A force graph rises linearly from 0 to 6 N over 2 m, then stays at 6 N for another 1 m. Find total work and explain your area decomposition.
Step 1. The first 2 m make a triangle of area ½(2 m)(6 N) = 6 J.
Why this step? The parallel force rises linearly from zero, so its average over this interval is 3 N rather than the final 6 N.
Step 2. The next 1 m make a rectangle of area (1 m)(6 N) = 6 J.
Why this step? Here the force is constant, so every part of the displacement has the same 6 N graph height.
Step 3. Total work is 6 + 6 = 12 J.
Why this step? The two nonoverlapping intervals both lie above the axis, so their positive work contributions add without cancellation.
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Question 6
Compare gravitational work on a book lifted to height h by two different paths. Then compare friction work for two rough paths of different lengths that share endpoints.
Step 1. For either lift from the same starting height to height h above it, gravitational work is W_g = −mgh for constant g.
Why this step? Gravity points downward, and its conservative work depends only on the common upward endpoint change, not intermediate turns.
Step 2. A longer rough route with constant friction magnitude has W_f = −f_kL, with larger |W_f| when L is larger.
Why this step? Sliding friction opposes every traveled segment, so additional path length removes additional mechanical energy even between the same endpoints.
Step 3. Equal gravitational work therefore does not imply equal friction work for the two paths.
Why this step? Conservative and nonconservative interactions use different dependencies; substituting endpoint height for friction path length would confuse them.