Apply Work-Energy Theorem - Grade 11 Physics

1. A $5.0\,\text{kg}$ crate moves up a $30^\circ$ incline with an initial speed of $2.0\,\text{m/s}$. A constant $60\,\text{N}$ force parallel to the incline pushes the crate uphill through $4.0\,\text{m}$. Friction of magnitude $5.0\,\text{N}$ acts downhill. Using $g=10\,\text{m/s}^2$, what is the crate's speed after the displacement?
   Answer: $7.2\,\text{m/s}$ Whenever you see a problem involving forces, motion along an incline, and asking for a final speed, reach for the Work-Energy Theorem: the net work done on an object equals its change in kinetic energy, $W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. First, identify every force acting along the displacement ($d = 4.0\,\text{m}$). The applied force does positive work: $W_{applied} = 60 \times 4.0 = 240\,\text{J}$. Gravity has a component pulling down the incline: $mg\sin30° = 5.0 \times 10 \times 0.5 = 25\,\text{N}$, doing negative work $W_{gravity} = -25 \times 4.0 = -100\,\text{J}$. Friction acts downhill too: $W_{friction} = -5.0 \times 4.0 = -20\,\text{J}$. Net work: $W_{net} = 240 - 100 - 20 = 120\,\text{J}$. Now solve for $v_f$: $120 = \frac{1}{2}(5.0)v_f^2 - \frac{1}{2}(5.0)(2.0)^2 = 2.5v_f^2 - 10$, so $2.5v_f^2 = 130$, giving $v_f = \sqrt{52} \approx 7.2\,\text{m/s}$. That confirms A. Choice B ($7.7\,\text{m/s}$) results from forgetting friction's contribution to net work. Choice C ($9.6\,\text{m/s}$) comes from ignoring gravity's component along the incline entirely. Choice D ($11.5\,\text{m/s}$) corresponds to using only the applied force and neglecting both friction and gravity — a common trap when students forget to account for all opposing forces. Study tip: Always list every force acting along the direction of motion before calculating net work. On incline problems, gravity's component ($mg\sin\theta$) is easy to overlook — it's not the full weight.
2. A $10\,\text{kg}$ sled is moving at $3.0\,\text{m/s}$ on a horizontal surface. A rope pulls the sled through $8.0\,\text{m}$ with a constant force of $50\,\text{N}$ directed $37^\circ$ above the horizontal. A constant friction force of $20\,\text{N}$ opposes the motion. Using $\cos 37^\circ=0.80$, what is the sled's final speed?
   Answer: $6.4\,\text{m/s}$ When a force is applied at an angle, only its horizontal component does work on the sled — this is a classic work-energy theorem problem. Your strategy: find the net work done on the sled, then use $W_{net} = \Delta KE$ to find the final speed. Start by identifying each work contribution over $d = 8.0\,\text{m}$. The rope's horizontal component is $F\cos37° = 50 \times 0.80 = 40\,\text{N}$, so the rope does $W_{rope} = 40 \times 8.0 = 320\,\text{J}$. Friction opposes motion, so $W_{friction} = -20 \times 8.0 = -160\,\text{J}$. Net work: $W_{net} = 320 - 160 = 160\,\text{J}$. Now apply the work-energy theorem. The sled's initial KE is $\frac{1}{2}(10)(3.0)^2 = 45\,\text{J}$, so final KE $= 45 + 160 = 205\,\text{J}$. Solving: $v_f = \sqrt{\frac{2 \times 205}{10}} = \sqrt{41} \approx 6.4\,\text{m/s}$. That confirms B. Choice A ($5.7\,\text{m/s}$) likely comes from forgetting the sled's initial kinetic energy entirely, starting from rest. Choice C ($7.5\,\text{m/s}$) results from using the full $50\,\text{N}$ instead of just the horizontal component — a very common trap when a force is angled. Choice D ($8.5\,\text{m/s}$) comes from ignoring friction altogether while also using the full force. Study tip: Whenever a force is applied at an angle, always decompose it first. Only the component parallel to displacement does work — writing $W = Fd\cos\theta$ before plugging in numbers keeps you from falling into the angle trap.
3. An object is acted on by several forces while moving along a curved path. The total work done by all forces over the complete path is zero. Which statement must be true?
   Answer: The object's final speed equals its initial speed, though the direction of motion may have changed. Whenever you see a question linking total work to motion, your first instinct should be the Work-Energy Theorem: the net work done on an object equals its change in kinetic energy. $W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ If the total work done by all forces is zero, then $\Delta KE = 0$, which means $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2$. Since mass is constant, this tells you directly that $v_f = v_i$ — the final speed equals the initial speed. That's exactly what D states, and it's the correct answer. Here's why the other choices fail. A is wrong because individual forces can each do nonzero work as long as their contributions cancel out in total — for example, friction doing negative work while a push does equal positive work. Forces don't need to cancel at every point. B confuses zero work with zero net force. A net force can exist and still do zero net work over a path (imagine a force always perpendicular to motion, like circular motion). Zero net work does not imply zero net force or constant velocity. C is the trickiest distractor — it overclaims. Equal speeds don't require equal velocity vectors. The object could be moving in a completely different direction at the end; the Work-Energy Theorem says nothing about direction, only magnitude of velocity (speed). Study tip: Always distinguish between speed (scalar, magnitude only) and velocity (vector, magnitude + direction). The Work-Energy Theorem governs kinetic energy, which depends only on speed — never direction.
4. A $2.0\,\text{kg}$ cart initially moves at $4.0\,\text{m/s}$ along a straight track. During the first $3.0\,\text{m}$, the net force in the direction of motion increases uniformly from $0\,\text{N}$ to $12\,\text{N}$. During the next $4.0\,\text{m}$, the net force is constant at $3.0\,\text{N}$ opposite the motion. What is the cart's final speed?
   Answer: $4.7\,\text{m/s}$ When a force varies with position (not time), the work-energy theorem is your most powerful tool. Instead of trying to find acceleration, ask: how much work does each force do, and how does that change kinetic energy? The work-energy theorem states $W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. Segment 1: The force increases uniformly from $0\,\text{N}$ to $12\,\text{N}$ over $3.0\,\text{m}$. For a linearly varying force, work equals the area of the triangle under the F-vs-x graph: $W_1 = \frac{1}{2}(12)(3.0) = 18\,\text{J}$. Segment 2: A constant $3.0\,\text{N}$ opposing force over $4.0\,\text{m}$ does negative work: $W_2 = -(3.0)(4.0) = -12\,\text{J}$. Total work: $W_{net} = 18 + (-12) = 6\,\text{J}$. Now apply the theorem: $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{net} = \frac{1}{2}(2.0)(4.0)^2 + 6 = 16 + 6 = 22\,\text{J}$, giving $v_f = \sqrt{\frac{2(22)}{2.0}} \approx 4.7\,\text{m/s}$, confirming A. Choice B (5.8 m/s) likely comes from using the full rectangle $12 \times 3 = 36\,\text{J}$ instead of the triangular area for segment 1. Choice C (6.3 m/s) results from ignoring the opposing force entirely. Choice D (6.6 m/s) likely adds both work values as positive, missing the sign of the opposing force. Study tip: Whenever a force changes with position, draw the F-vs-x graph and calculate work as the geometric area — triangles, rectangles, or trapezoids. This sidesteps complex kinematics entirely.
5. A $500\,\text{kg}$ roller-coaster car moves at $10\,\text{m/s}$ at a height of $20\,\text{m}$. It later reaches a height of $12\,\text{m}$. Between these points, a drive system does $20\,\text{kJ}$ of positive work, and resistive forces do $35\,\text{kJ}$ of negative work. Using $g=10\,\text{m/s}^2$, what is the car's speed at the lower point?
   Answer: $14.1\,\text{m/s}$ Whenever you see a roller-coaster problem with external forces (drives, friction, air resistance), reach for the Work-Energy Theorem in its full form: the net work done on an object equals its change in kinetic energy. Here, that means accounting for gravitational potential energy changes and the extra work inputs and losses. Set up the energy equation: $KE_1 + PE_1 + W_{\text{drive}} + W_{\text{resist}} = KE_2 + PE_2$ Plugging in values (with $m = 500\,\text{kg}$, $v_1 = 10\,\text{m/s}$, $h_1 = 20\,\text{m}$, $h_2 = 12\,\text{m}$): $KE_1 = \tfrac{1}{2}(500)(10)^2 = 25{,}000\,\text{J}$ $PE_1 = (500)(10)(20) = 100{,}000\,\text{J}$ $PE_2 = (500)(10)(12) = 60{,}000\,\text{J}$ Total energy available at point 2: $25{,}000 + 100{,}000 + 20{,}000 - 35{,}000 = 110{,}000\,\text{J}$ Subtract $PE_2$: $KE_2 = 110{,}000 - 60{,}000 = 50{,}000\,\text{J}$ Solving: $v_2 = \sqrt{\frac{2(50{,}000)}{500}} = \sqrt{200} \approx 14.1\,\text{m/s}$ — confirming C. Choice A (6.3 m/s) likely comes from forgetting to include the gravitational PE drop (the height decrease adds energy). Choice B (11.0 m/s) results from ignoring all external work entirely and treating it as a pure conservation problem, which gives only a small speed increase. Choice D (16.1 m/s) comes from adding the resistive work instead of subtracting it, flipping its sign. Your strategy: always assign signs carefully — positive work adds energy, negative work removes it. Write the full energy balance before calculating anything.
6. A car of mass $m$ traveling at $20\,\text{m/s}$ is brought to rest by a constant braking force over a distance of $40\,\text{m}$. A second car of mass $0.80m$ is stopped by a braking force of the same magnitude while traveling initially at $30\,\text{m/s}$. How far does the second car travel before stopping?
   Answer: $72\,\text{m}$ Whenever you see a problem involving a braking force stopping a moving car, reach for the work-energy theorem: the net work done on an object equals its change in kinetic energy. Here, the braking force does negative work over the stopping distance, so $F \cdot d = \frac{1}{2}mv^2$. For the first car: $F \cdot 40 = \frac{1}{2}m(20)^2 = 200m$, which gives you $F = 5m$. Now apply the same relationship to the second car, using mass $0.80m$, initial speed $30\,\text{m/s}$, and the same braking force $F = 5m$: $F \cdot d = \frac{1}{2}(0.80m)(30)^2$ $5m \cdot d = \frac{1}{2}(0.80m)(900) = 360m$ $d = \frac{360m}{5m} = 72\,\text{m}$ That confirms C is correct. Looking at the wrong choices: A (50 m) likely comes from simply scaling by the ratio of speeds $(30/20) \times 40 = 60\,\text{m})$ — but even that gives 60 m, not 50, so A doesn't follow from any clean logic. B (60 m) is the trap of scaling stopping distance linearly with speed, ignoring that kinetic energy depends on $v^2$. D (90 m) may come from forgetting to account for the reduced mass of the second car — if you use mass $m$ instead of $0.80m$, you get $d = 90\,\text{m}$. Study tip: Stopping distance scales with $v^2$, not $v$. On any braking or collision problem, write out $Fd = \frac{1}{2}mv^2$ immediately — it keeps all variables visible and prevents shortcuts that lead to wrong answers.
7. A cable lifts a $600\,\text{kg}$ elevator upward through $5.0\,\text{m}$. During the lift, the elevator's speed increases from $2.0\,\text{m/s}$ to $6.0\,\text{m/s}$. Neglect friction and use $g=10\,\text{m/s}^2$. How much work does the cable do on the elevator?
   Answer: $39.6\,\text{kJ}$ When a question involves a force moving an object that also changes speed, you need the Work-Energy Theorem: the net work done on an object equals its change in kinetic energy. But here's the key insight — the cable isn't the only force acting. Gravity also does work (negative work, since it opposes the upward motion). So the cable must do work against gravity AND provide the kinetic energy increase. The work done by the cable equals the work done against gravity plus the gain in kinetic energy: $W_\text{cable} = W_\text{gravity} + \Delta KE$ Work against gravity: $W_g = mgh = 600 \times 10 \times 5.0 = 30{,}000\,\text{J}$ Change in kinetic energy: $\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(600)(6^2 - 2^2) = 300(36-4) = 9{,}600\,\text{J}$ Total: $W_\text{cable} = 30{,}000 + 9{,}600 = 39{,}600\,\text{J} = \mathbf{39.6\,\text{kJ}}$, confirming D. Choice A ($9.6\,\text{kJ}$) only accounts for the kinetic energy change — it forgets that the cable also had to lift the elevator against gravity. Choice B ($20.4\,\text{kJ}$) has no clean physical meaning and likely results from arithmetic errors mixing up the two components. Choice C ($30.0\,\text{kJ}$) only includes the gravitational work term — it ignores the acceleration entirely, as if the elevator moved at constant speed. Your study tip: whenever an object is both rising and accelerating, the cable (or applied force) must do work for both effects. Always split the problem into $W_\text{gravity} + \Delta KE$ — don't let one cancel the other out in your head.
8. A $4.0\,\text{kg}$ box slides horizontally at $8.0\,\text{m/s}$. A constant applied force of $6.0\,\text{N}$ acts in the direction of motion while a constant resistive force acts opposite the motion. After the box travels $10\,\text{m}$, its speed is $5.0\,\text{m/s}$. What is the magnitude of the resistive force?
   Answer: $13.8\,\text{N}$ When a problem gives you initial speed, final speed, displacement, and forces, the work-energy theorem is your go-to framework: the net work done on an object equals its change in kinetic energy, $W_{net} = \Delta KE$. Start by calculating the change in kinetic energy: $\Delta KE = \frac{1}{2}(4.0)(5.0)^2 - \frac{1}{2}(4.0)(8.0)^2 = 50 - 128 = -78\,\text{J}$. The negative sign tells you the box slowed down — the resistive force won out overall. Now set up the net work equation. The applied force does positive work and the resistive force does negative work over the 10 m displacement: $W_{net} = (F_{applied} - F_{resistive}) \cdot d = \Delta KE$ $(6.0 - F_r)(10) = -78$ $6.0 - F_r = -7.8$ $F_r = 13.8\,\text{N}$ That confirms C is correct. Now for the traps: A (1.8 N) comes from forgetting the negative sign on $\Delta KE$ and solving $(6.0 - F_r)(10) = +78$, giving $F_r = -1.8\,\text{N}$ — a sign error leading to a nonsensical result that students round into 1.8 N. B (7.8 N) results from ignoring the applied force entirely and setting $F_r \cdot 10 = 78$, forgetting that net work involves all forces. D (19.8 N) comes from adding the applied force instead of subtracting it when solving, a classic algebra sign mistake. Your study tip: always write the full net work equation before solving. Listing every force — with correct signs — before plugging in numbers prevents nearly all errors on work-energy problems.
9. A $2.0\,\text{kg}$ block is initially at rest against a spring compressed by $0.30\,\text{m}$ on a horizontal surface. The spring constant is $400\,\text{N/m}$. After release, the block moves through the compression distance and then travels an additional $0.50\,\text{m}$ beyond the spring's natural length. A constant friction force of $6.0\,\text{N}$ acts throughout the entire motion. What is the block's speed at that point?
   Answer: $3.6\,\text{m/s}$ Whenever you see a spring-release problem with friction, your go-to tool is the work-energy theorem: the net work done on the block equals its change in kinetic energy. The spring does positive work, while friction does negative work over the entire distance traveled. The total distance the block travels is the compression distance plus the extra distance: $d = 0.30 + 0.50 = 0.80\,\text{m}$. The spring's elastic potential energy converts to kinetic energy, minus the energy lost to friction: $\frac{1}{2}kx^2 - f_k d = \frac{1}{2}mv^2$ Plugging in values: $\frac{1}{2}(400)(0.30)^2 - (6.0)(0.80) = \frac{1}{2}(2.0)v^2$ $18 - 4.8 = v^2$ $13.2 = v^2 \implies v = \sqrt{13.2} \approx 3.6\,\text{m/s}$ This confirms A is correct. Choice B (3.9 m/s) results from using only the extra 0.50 m for friction's work, forgetting that friction acts during the compression phase too — the block moves 0.80 m total, not 0.50 m. Choice C (4.2 m/s) comes from ignoring friction entirely, treating it as a frictionless problem. Choice D (5.6 m/s) results from forgetting to take the square root of $v^2$, a classic arithmetic slip under pressure. Your study tip: always sketch the full path and ask "over what total distance does friction act?" Friction is sneaky — students routinely forget to count the compression distance, which is where choice B trips you up.

Start this quiz