Work, Energy and Power Practice Problems (AP Physics 1)

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Work done by a force along the motion

   A worker pushes a 24 kg crate 7.5 m in a straight line across a level floor with a constant 62 N horizontal force. Find the work the worker does on the crate, and the work done on it by gravity and by the normal force.

   Show the worked solution

   1. The push is horizontal and the displacement is horizontal, so the angle between them is zero and $\cos 0^\circ = 1$. The whole force is parallel to the motion.

      $$W = Fd\cos\theta = (62\ \text{N})(7.5\ \text{m})(1) = 465\ \text{J}$$

   2. Gravity points straight down and the displacement is horizontal, so the angle between them is $90^\circ$ and $\cos 90^\circ = 0$.

      $$W_{\text{gravity}} = (235.2\ \text{N})(7.5\ \text{m})\cos 90^\circ = 0$$

   3. The normal force points straight up, also perpendicular to the motion, so it does zero work too. Neither of them transfers any energy to the crate however far it travels.

   4. The 24 kg was only needed to state the weight in that middle step, and it cancelled out of the answer anyway. Work asks for force and displacement, not mass.

   The worker does $465\ \text{J}$ of work, and gravity and the normal force each do zero. If the crate were dragged back over the same 7.5 m while the worker kept pushing the original way, that same push would do $-465\ \text{J}$: the force did not change, the displacement reversed, and the angle between them went from $0^\circ$ to $180^\circ$.

2. 2. Work at an angle, and a negative-work case

   A 19 kg suitcase is pulled 11 m in a straight line along a level floor by a strap held at $32^\circ$ above the horizontal, with a constant tension of 45 N. A constant 13 N friction force opposes the motion. Find the work done on the suitcase by the strap, by friction, by gravity and by the normal force, then the net work.

   Show the worked solution

   1. Take positive $x$ as the direction of travel. The strap makes $32^\circ$ with the displacement, so only its horizontal component transfers energy.

      $$F_{\parallel} = (45\ \text{N})\cos 32^\circ = (45)(0.84805) = 38.16\ \text{N}$$
      $$W_{\text{strap}} = F_{\parallel} d = (38.16\ \text{N})(11\ \text{m}) = 419.8\ \text{J}$$

   2. Friction points backwards along the floor, directly opposite the displacement, so the angle is $180^\circ$ and $\cos 180^\circ = -1$.

      $$W_{\text{friction}} = (13\ \text{N})(11\ \text{m})(-1) = -143\ \text{J}$$

      That negative sign is not a bookkeeping convention. It says 143 J of energy left the suitcase's kinetic store.

   3. Gravity and the normal force are both vertical while the displacement is horizontal, so both do zero work: $\cos 90^\circ = 0$. The suitcase's 186.2 N weight never appears in the answer.

   4. Net work is the scalar sum of all four.

      $$W_{\text{net}} = 419.8\ \text{J} - 143\ \text{J} + 0 + 0 = 276.8\ \text{J}$$

   5. Two significant figures: about $4.2 \times 10^2\ \text{J}$ from the strap, $-1.4 \times 10^2\ \text{J}$ from friction, and $2.8 \times 10^2\ \text{J}$ net.

   6. Worth noticing what the strap's vertical component did. It is $(45)(\sin 32^\circ) = 23.85\ \text{N}$ upward, and it does no work, because it is perpendicular to the motion. It still matters physically, since it reduces the normal force and therefore the friction, but it transfers no energy.

   $W_{\text{strap}} = +4.2 \times 10^2\ \text{J}$, $W_{\text{friction}} = -1.4 \times 10^2\ \text{J}$, $W_{\text{gravity}} = 0$, $W_{\text{normal}} = 0$, and $W_{\text{net}} = +2.8 \times 10^2\ \text{J}$. Three different angles between force and displacement give three different signs: $32^\circ$ gives positive work, $180^\circ$ gives negative, and $90^\circ$ gives none.

3. 3. The work-energy theorem instead of kinematics

   A 1450 kg car speeds up from 18 m/s to 27 m/s along a straight level road, covering 140 m while it does so. Find the net work done on the car and the average net force acting on it over that stretch.

   Show the worked solution

   1. No time is given and none is asked for, which is the signal to use energy rather than a kinematic chain. The work-energy theorem says the net work equals the change in kinetic energy.

      $$W_{\text{net}} = \Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \tfrac{1}{2}m\left(v_f^2 - v_i^2\right)$$

   2. Factoring the mass out first keeps the arithmetic small.

      $$v_f^2 - v_i^2 = (27\ \text{m/s})^2 - (18\ \text{m/s})^2 = 729 - 324 = 405\ \text{m}^2/\text{s}^2$$
      $$W_{\text{net}} = \tfrac{1}{2}(1450\ \text{kg})(405\ \text{m}^2/\text{s}^2) = (725)(405) = 293{,}625\ \text{J}$$

   3. The net force is constant only on average here, but average force times distance still gives the work, so invert that.

      $$F_{\text{net}} = \frac{W_{\text{net}}}{d} = \frac{293{,}625\ \text{J}}{140\ \text{m}} = 2097\ \text{N}$$

   4. Two significant figures: $W_{\text{net}} = 2.9 \times 10^5\ \text{J}$ and $F_{\text{net}} = 2.1 \times 10^3\ \text{N}$.

   5. Cross-check by the longer route. Constant acceleration over 140 m would need $a = (v_f^2 - v_i^2)/(2d) = 405/280 = 1.446\ \text{m/s}^2$, and $F = ma = (1450)(1.446) = 2097\ \text{N}$. Identical, because $v^2 = v_0^2 + 2a\Delta x$ is the work-energy theorem in disguise: multiply it through by $\tfrac{1}{2}m$ and you get $\Delta K = F d$.

   $W_{\text{net}} = 2.9 \times 10^5\ \text{J}$ and $F_{\text{net}} = 2.1 \times 10^3\ \text{N}$ forwards. The energy route skipped acceleration entirely. It also survives a stretch of road where the force is not constant, which the kinematic equations do not, since only the average matters to the work.

4. 4. Gravitational potential energy, and why the zero does not matter

   A 2.6 kg book is lifted from the floor onto a shelf 1.9 m above it, then later moved down to a second shelf 0.8 m above the floor. Find the change in gravitational potential energy of the book-Earth system for each move and for the two moves combined. Then state what changes if you had put the zero of potential energy at the upper shelf instead of the floor.

   Show the worked solution

   1. The sheet prints the relationship as a change, $\Delta U_g = mg\Delta y$, and that is the safest way to use it. Compute the weight once.

      $$mg = (2.6\ \text{kg})(9.8\ \text{m/s}^2) = 25.48\ \text{N}$$

   2. First move, floor to the 1.9 m shelf, so $\Delta y = +1.9\ \text{m}$.

      $$\Delta U_1 = (25.48\ \text{N})(+1.9\ \text{m}) = +48.41\ \text{J}$$

   3. Second move, 1.9 m down to 0.8 m, so $\Delta y = 0.8 - 1.9 = -1.1\ \text{m}$. The sign of $\Delta y$ carries straight through.

      $$\Delta U_2 = (25.48\ \text{N})(-1.1\ \text{m}) = -28.03\ \text{J}$$

   4. Combined, either by adding the two or by using the overall displacement $\Delta y = +0.8\ \text{m}$ in one step.

      $$\Delta U_{\text{total}} = 48.41 - 28.03 = 20.38\ \text{J} \qquad \text{and} \qquad (25.48)(0.8) = 20.38\ \text{J}$$

      The two agree because gravity is a conservative force: the work it does is path-independent and depends only on the starting and finishing configurations (essential knowledge 3.2.A.1.i).

   5. Moving the zero to the upper shelf changes every individual value of $U_g$ by the same constant, $-48.41\ \text{J}$. Every difference is unaffected, so all three answers above stay exactly as they are. Only differences ever appear in the physics, which is why the sheet prints $\Delta U_g$ and not $U_g$.

   6. Two significant figures: $+48\ \text{J}$, $-28\ \text{J}$, $+20\ \text{J}$.

   $\Delta U_1 = +48\ \text{J}$, $\Delta U_2 = -28\ \text{J}$, and $\Delta U_{\text{total}} = +20\ \text{J}$. Relocating the zero shifts every stored value by a fixed amount and changes none of the three answers. If a solution of yours ever depends on where you put the zero, you have used an absolute $U_g$ somewhere a difference belonged.

5. 5. Elastic potential energy in a spring

   A spring with force constant $k = 180\ \text{N/m}$ is compressed 0.24 m from its natural length. (a) Find the elastic potential energy stored in it. (b) How far would you have to compress the same spring to store twice that energy?

   Show the worked solution

   1. Part (a). The sheet prints $U_s = \tfrac{1}{2}k(\Delta x)^2$, where $\Delta x$ is the displacement from the natural length, not the length of the spring.

      $$U_s = \tfrac{1}{2}(180\ \text{N/m})(0.24\ \text{m})^2 = (90)(0.0576) = 5.184\ \text{J}$$

   2. Part (b). The energy goes with the square of the compression, so doubling the energy does not need double the compression. Set up the ratio rather than recomputing from scratch.

      $$\frac{U_2}{U_1} = \frac{(\Delta x_2)^2}{(\Delta x_1)^2} = 2 \quad \Rightarrow \quad \Delta x_2 = \Delta x_1 \sqrt{2}$$

   3. Substitute the numbers.

      $$\Delta x_2 = (0.24\ \text{m})(1.4142) = 0.3394\ \text{m}$$

   4. Check by going forwards: $\tfrac{1}{2}(180)(0.3394)^2 = (90)(0.11519) = 10.37\ \text{J}$, which is twice 5.184 J to the digits shown.

   5. Two significant figures: $U_s = 5.2\ \text{J}$ and $\Delta x_2 = 0.34\ \text{m}$. Doubling the compression instead, to 0.48 m, would have stored four times the energy, $20.7\ \text{J}$.

   (a) $U_s = 5.2\ \text{J}$. (b) $0.34\ \text{m}$, which is $\sqrt{2}$ times the original compression rather than twice it. That squared dependence is why the last centimetre of compression on a stiff spring stores far more than the first.

6. 6. Conservation of energy on a frictionless track

   A 0.42 kg cart is released from rest at the top of a frictionless track, 1.35 m above the bottom. Find its speed at the bottom of the track, and its speed as it passes a point 0.55 m above the bottom. State whether the shape of the track between those points matters.

   Show the worked solution

   1. Take the cart and the Earth as the system. The track is frictionless and the normal force does no work on the cart (it is perpendicular to the motion everywhere), so no work crosses the system boundary and the mechanical energy is constant. That is essential knowledge 3.4.C.2.

      $$K_i + U_i = K_f + U_f$$

   2. The cart starts from rest, so $K_i = 0$. Measure heights from the bottom of the track, so $U_f = 0$ there.

      $$mgh = \tfrac{1}{2}mv^2$$

   3. The mass appears on both sides and cancels, so the answer is the same for any cart.

      $$v = \sqrt{2gh} = \sqrt{2(9.8\ \text{m/s}^2)(1.35\ \text{m})} = \sqrt{26.46} = 5.144\ \text{m/s}$$

   4. For the point 0.55 m up, only the drop so far has been converted, which is $1.35 - 0.55 = 0.80\ \text{m}$.

      $$v = \sqrt{2(9.8\ \text{m/s}^2)(0.80\ \text{m})} = \sqrt{15.68} = 3.960\ \text{m/s}$$

   5. The shape of the track between the two points is irrelevant. Gravity is conservative, so the work it does depends only on the vertical drop, and the normal force from the track does no work at any point along it however the track curves. The cart arrives with the same speed whether the drop is a straight ramp, a curve, or a loop.

   6. Significant figures: the 1.35 m carries three, giving $5.14\ \text{m/s}$; the 0.80 m drop carries two, giving $4.0\ \text{m/s}$.

   $v = 5.14\ \text{m/s}$ at the bottom and $4.0\ \text{m/s}$ at the 0.55 m point. The 0.42 kg cancels and never affects either speed. The shape of the track does not matter, only the height dropped, which is the practical payoff of gravity being a conservative force.

7. 7. A spring launcher: two stores in one motion

   A spring with force constant $k = 260\ \text{N/m}$ is compressed 0.18 m against a 0.75 kg cart resting on a frictionless horizontal track. The cart is released, leaves the spring at the spring's natural length, and then runs onto a frictionless ramp that curves upward. Find the cart's speed as it leaves the spring, and the maximum height it reaches on the ramp.

   Show the worked solution

   1. Take the cart, spring, track and Earth as the system. Nothing dissipates energy and nothing does work from outside, so the total mechanical energy is fixed at whatever the spring started with.

      $$U_s = \tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(260\ \text{N/m})(0.18\ \text{m})^2 = (130)(0.0324) = 4.212\ \text{J}$$

   2. First handoff: elastic to kinetic. At the moment the cart separates from the spring, the spring is at its natural length so $U_s = 0$, and the track is still horizontal so $\Delta U_g = 0$.

      $$\tfrac{1}{2}mv^2 = 4.212\ \text{J}$$
      $$v = \sqrt{\frac{2(4.212\ \text{J})}{0.75\ \text{kg}}} = \sqrt{11.232} = 3.351\ \text{m/s}$$

   3. Second handoff: kinetic to gravitational. At the highest point the cart is momentarily at rest, so all 4.212 J now sits in the gravitational store.

      $$mgh = 4.212\ \text{J}$$

   4. Solve for the height.

      $$h = \frac{4.212\ \text{J}}{(0.75\ \text{kg})(9.8\ \text{m/s}^2)} = \frac{4.212}{7.35} = 0.5731\ \text{m}$$

   5. Check by the other route: $h = v^2/(2g) = 11.232/19.6 = 0.5731\ \text{m}$. The two agree, as they must, since the middle step was just the same energy written differently.

   6. Two significant figures: $U_s = 4.2\ \text{J}$, $v = 3.4\ \text{m/s}$, $h = 0.57\ \text{m}$.

   The cart leaves the spring at $3.4\ \text{m/s}$ and climbs to $0.57\ \text{m}$ above the track. The same 4.2 J appears in three forms in turn, and you can jump straight from the first to the third if the middle one is not asked for. The ramp's angle never entered the calculation.

8. 8. A rough ramp, where mechanical energy is not conserved

   A 3.2 kg box is released from rest at the top of a ramp and slides down to the bottom, dropping 1.6 m in height along a 4.0 m path on the ramp surface. It arrives at the bottom moving at 4.1 m/s. Find how much mechanical energy was dissipated, the average friction force on the box, and the speed it would have reached on the same ramp with no friction.

   Show the worked solution

   1. Take the box, ramp and Earth as the system, and set the zero of gravitational potential energy at the bottom of the ramp. The box starts at rest, so all its initial mechanical energy is gravitational.

      $$U_i = mgh = (3.2\ \text{kg})(9.8\ \text{m/s}^2)(1.6\ \text{m}) = 50.18\ \text{J}$$

   2. At the bottom all of the remaining mechanical energy is kinetic.

      $$K_f = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(3.2\ \text{kg})(4.1\ \text{m/s})^2 = (1.6)(16.81) = 26.90\ \text{J}$$

   3. The gap is the mechanical energy that left the mechanical stores. It did not vanish: the Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. Total energy is still conserved (3.4.C.1).

      $$E_{\text{dissipated}} = U_i - K_f = 50.18\ \text{J} - 26.90\ \text{J} = 23.28\ \text{J}$$

   4. To get the friction force, use the CED's working rule from 3.2.A.4.iii: the energy dissipated by friction is equated to the friction force times the length of the path over which it acts. The path along the ramp surface is 4.0 m, not the 1.6 m height.

      $$F_f = \frac{E_{\text{dissipated}}}{d} = \frac{23.28\ \text{J}}{4.0\ \text{m}} = 5.82\ \text{N}$$

   5. Frictionless comparison. With nothing dissipated, all 50.18 J would arrive as kinetic energy and the mass would cancel.

      $$v = \sqrt{2gh} = \sqrt{2(9.8)(1.6)} = \sqrt{31.36} = 5.6\ \text{m/s}$$

   6. Two significant figures: $23\ \text{J}$ dissipated, $F_f = 5.8\ \text{N}$, and a frictionless arrival speed of $5.6\ \text{m/s}$ against the actual 4.1 m/s.

   $23\ \text{J}$ of mechanical energy was dissipated, the average friction force was $5.8\ \text{N}$, and a frictionless ramp would have delivered the box at $5.6\ \text{m/s}$ rather than 4.1 m/s. Friction removed 46 percent of the kinetic energy the box would otherwise have arrived with, yet cost it only 1.5 m/s of speed, because kinetic energy goes with the square of the speed.

9. 9. Average power on a climb

   A 68 kg firefighter climbs a vertical ladder 8.4 m high at a steady pace, taking 12 s. (a) Find the work done against gravity and the average power developed. (b) A second firefighter of the same mass climbs the same ladder in 18 s. Find the work and the average power for that climb, and say which quantity changed.

   Show the worked solution

   1. Part (a). Climbing at a steady pace means the kinetic energy is unchanged from start to finish, so all of the work goes into the gravitational store.

      $$W = mg\Delta y = (68\ \text{kg})(9.8\ \text{m/s}^2)(8.4\ \text{m}) = 5598\ \text{J}$$

   2. Average power is that energy divided by the time it took, which is exactly what the sheet prints as $P_{\text{avg}} = W/\Delta t$.

      $$P_{\text{avg}} = \frac{5598\ \text{J}}{12\ \text{s}} = 466.5\ \text{W}$$

   3. Part (b). Same mass, same height, so the work is identical at 5598 J. Only the denominator changed.

      $$P_{\text{avg}} = \frac{5598\ \text{J}}{18\ \text{s}} = 311.0\ \text{W}$$

   4. Two significant figures: $W = 5.6 \times 10^3\ \text{J}$ in both cases, with $P_{\text{avg}} = 4.7 \times 10^2\ \text{W}$ for the faster climb and $3.1 \times 10^2\ \text{W}$ for the slower one.

   5. The ratio of the powers, $466.5/311.0 = 1.50$, is exactly the inverse ratio of the times, $18/12 = 1.50$. Power is a rate, so at fixed work it is inversely proportional to time.

   (a) $W = 5.6 \times 10^3\ \text{J}$ and $P_{\text{avg}} = 4.7 \times 10^2\ \text{W}$. (b) The same $5.6 \times 10^3\ \text{J}$ of work, but only $3.1 \times 10^2\ \text{W}$. Work counts the energy moved; power counts how fast you moved it. Both firefighters did the same job and only one of them did it quickly.

10. 10. Instantaneous power from a force at an angle

    A tow truck pulls a broken-down car along a level road at a constant 12 m/s. The tow cable is straight and makes an angle of $22^\circ$ with the road, and the tension in it is 1900 N. (a) Find the instantaneous power the cable delivers to the car. (b) Find the work the cable does in 30 s of towing. (c) State the net power delivered to the car, and explain the value.

    Show the worked solution

    1. Part (a). The sheet prints $P_{\text{inst}} = F_{\parallel} v = F v \cos\theta$, with $\theta$ the angle between the force and the velocity. Only the component of the tension along the road transfers energy.

       $$F_{\parallel} = (1900\ \text{N})\cos 22^\circ = (1900)(0.92718) = 1762\ \text{N}$$
       $$P_{\text{inst}} = (1762\ \text{N})(12\ \text{m/s}) = 21{,}140\ \text{W}$$

    2. Part (b). Both the force and the speed are constant, so the instantaneous power never changes and the energy transferred is just power times time.

       $$W = P\,\Delta t = (21{,}140\ \text{W})(30\ \text{s}) = 6.34 \times 10^5\ \text{J}$$

    3. Check that against the work formula directly. In 30 s the car covers $(12\ \text{m/s})(30\ \text{s}) = 360\ \text{m}$, so

       $$W = F_{\parallel} d = (1762\ \text{N})(360\ \text{m}) = 6.34 \times 10^5\ \text{J}$$

       The two agree. Because $F$ and $v$ are both constant here, the average power over the 30 s equals the instantaneous power at every moment inside it, which is not true in general.

    4. Part (c). The car moves at constant speed, so its kinetic energy is not changing, so by the work-energy theorem the net work on it is zero and the net power is zero. The cable delivers about 21 kW and drag, rolling resistance and friction remove about 21 kW at the same time.

    5. Two significant figures: $P_{\text{inst}} = 2.1 \times 10^4\ \text{W}$, about 21 kW, and $W = 6.3 \times 10^5\ \text{J}$ over the 30 s.

    (a) $2.1 \times 10^4\ \text{W}$, about 21 kW. (b) $6.3 \times 10^5\ \text{J}$. (c) Zero net power, because the speed is constant and so the kinetic energy is not changing. Angling the cable at $22^\circ$ costs about 7 percent of the power the same tension would deliver along the road, since $\cos 22^\circ = 0.927$.

11. 11. Spring, rough patch and ramp in one motion

    A 1.6 kg block is pressed against a spring of force constant $k = 540\ \text{N/m}$, compressing it 0.12 m, on a horizontal surface. The block is released. It leaves the spring, crosses a rough patch 0.85 m long where the coefficient of kinetic friction is $\mu_k = 0.15$, and then runs onto a frictionless ramp. Find the block's speed when it leaves the rough patch and the maximum height it reaches on the ramp.

    Show the worked solution

    1. Track the energy from one end to the other. Start with what the spring stored.

       $$U_s = \tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(540\ \text{N/m})(0.12\ \text{m})^2 = (270)(0.0144) = 3.888\ \text{J}$$

    2. Now the rough patch. The surface is level and nothing presses on the block vertically except gravity, so the normal force is the full weight and the friction force is constant across the patch.

       $$F_f = \mu_k mg = (0.15)(1.6\ \text{kg})(9.8\ \text{m/s}^2) = (0.15)(15.68\ \text{N}) = 2.352\ \text{N}$$

    3. Apply the CED's rule from 3.2.A.4.iii, friction force times path length.

       $$E_{\text{dissipated}} = F_f d = (2.352\ \text{N})(0.85\ \text{m}) = 1.999\ \text{J}$$

       That is a little over half of everything the spring stored, so expect a modest answer.

    4. What survives the patch is kinetic energy, since the surface is still level and no height has been gained.

       $$K = U_s - E_{\text{dissipated}} = 3.888\ \text{J} - 1.999\ \text{J} = 1.889\ \text{J}$$
       $$v = \sqrt{\frac{2(1.889\ \text{J})}{1.6\ \text{kg}}} = \sqrt{2.361} = 1.537\ \text{m/s}$$

    5. The ramp is frictionless, so nothing more is lost and all 1.889 J converts to gravitational potential energy at the highest point, where the block is momentarily at rest.

       $$h = \frac{K}{mg} = \frac{1.889\ \text{J}}{(1.6\ \text{kg})(9.8\ \text{m/s}^2)} = \frac{1.889}{15.68} = 0.1205\ \text{m}$$

    6. Two significant figures: $v = 1.5\ \text{m/s}$ and $h = 0.12\ \text{m}$.

    7. Two traps to name. Do not apply $\mu_k$ to the ramp, which the problem says is frictionless. And do not use the ramp's vertical rise as a friction path length; friction acted only over the 0.85 m of rough level floor.

    The block leaves the rough patch at $1.5\ \text{m/s}$ and climbs to $0.12\ \text{m}$ on the ramp. Of the 3.9 J the spring stored, 2.0 J went to thermal energy in the rough patch and 1.9 J made it to the top of the climb. Removing the rough patch would have more than doubled the height, to $3.888/15.68 = 0.25\ \text{m}$.

12. 12. Explain why: a perpendicular force does no work

    A ball on a light string is whirled in a horizontal circle at constant speed on a frictionless table, and separately a crate slides in a straight line across a level floor. In the first case the string tension does no work on the ball; in the second the normal force does no work on the crate. Explain why in both cases, and say what a perpendicular force does change.

    Show the worked solution

    1. Start from what the sheet prints: $W = F_{\parallel}d = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement. At $\theta = 90^\circ$, $\cos 90^\circ = 0$, so the work is zero however large the force and however far the object travels. There is no exception to hunt for; the cosine is doing all of it.

    2. The ball. In uniform circular motion the velocity is tangent to the circle at every instant, and the tension points along the string towards the centre. Those two are perpendicular at every instant, so over any interval the tension does zero work and the ball's kinetic energy never changes. That is consistent with the given fact that the speed is constant.

    3. The crate. The floor's normal force points straight up and the displacement is horizontal. Perpendicular again, so zero work, no matter how heavy the crate. The same argument rules out gravity doing work on anything moving horizontally.

    4. The CED states both sides of this explicitly. Essential knowledge 3.2.A.3.i says only the component of the force parallel to the displacement of the point of application of the force will change the system's total energy. Essential knowledge 3.2.A.3.ii says the component perpendicular to the displacement of the system's centre of mass can change the direction of the system's motion without changing the system's kinetic energy.

    5. So a perpendicular force is not idle. It is the only thing turning the ball, and without it the ball would fly off along the tangent in a straight line. It changes the direction of the velocity while leaving its magnitude alone, which is precisely what happens in uniform circular motion.

    6. One consequence worth carrying forward: any force that is always perpendicular to the motion can be left out of an energy accounting entirely. That is why the normal force never appears in a conservation-of-energy equation for an object sliding on a track, however sharply the track curves.

    Both forces are perpendicular to the displacement, and $W = Fd\cos 90^\circ = 0$. Essential knowledge 3.2.A.3.i says only the parallel component changes the system's total energy, and 3.2.A.3.ii says the perpendicular component can change the direction of the motion without changing the kinetic energy. A perpendicular force steers rather than speeds up, which is why the string can hold a ball in a circle at unchanging speed and why the normal force can be dropped from any energy equation.