AP Physics 1 Unit 3: Work Energy and Power — Free Review Games.

W = Fd = 50 x 10 = 500 J.

KE = 1/2 mv^2 = 1/2(4)(25) = 50 J.

PE = mgh = 2(10)(10) = 200 J.

When only conservative forces act (no external work), total mechanical energy (KE + PE) is conserved.

Power is the rate of doing work: P = W/t, measured in watts.

PE = 1/2 kx^2 = 1/2(200)(0.01) = 1 J.

P = Fv = mgv. v = P/(mg) = 500/(100 x 10) = 0.5 m/s.

W = Fd*cos(theta) = 30(4)(cos60) = 30(4)(0.5) = 60 J.

Using energy conservation: $mgh = \frac{1}{2} mv^2$. \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \text{ m/s}\).

The work-energy theorem states W_net = delta KE = 1/2 mv_f^2 - 1/2 mv_i^2.

Energy at bottom: mgh = 2(10)(5) = 100 J. Friction force = mu_k*mg = 0.5(2)(10) = 10 N. Distance = Energy/Friction = 100/10 = 10 m.

1/2 kx^2 = mgh. 1/2(800)(0.0625) = 0.5(10)h. 25 = 5h. h = 5.0 m.

\(KE = \frac{1}{2} mv^2\). If KE doubles: $2(\frac{1}{2} mv^2) = \frac{1}{2} m(v_{new})^2$. $v_{new} = v\sqrt{2}$.

Work = delta KE = 0 - 1/2(1500)(20^2) = -1/2(1500)(400) = -300,000 J. Negative because friction opposes motion.

mgh - friction = 1/2 mv^2. 500(10)(40) - 20000 = 1/2(500)v^2. 200000 - 20000 = 250v^2. v^2 = 720. v = 26.8 m/s.

Work is defined as \(W = Fd\cos\theta\), and multiplying newtons by meters gives joules, the SI unit of energy transfer. The Watt is wrong because it measures the rate of energy transfer (power), not the energy itself. Remembering that work and energy share the same unit helps you connect the work-energy theorem to energy calculations on the exam.

Kinetic energy is derived from the work-energy theorem as \(KE = \frac{1}{2}mv^2\), since integrating \(F\,dx = ma\,dx\) over velocity yields this exact form. The choice $KE = mgh$ is wrong because that expression defines gravitational potential energy, not kinetic energy. Keep the factor of one-half in mind, since forgetting it is a very common exam error.

Gravitational potential energy near Earth's surface equals $mgh$ because the work needed to lift a mass a height \(h\) against a constant gravitational force \(mg\) is $mgh$. The expression \(\frac{1}{2}kx^2\) is wrong because that is elastic potential energy stored in a spring, not gravitational energy. Distinguishing energy formulas by the type of force involved is essential for setting up conservation of energy equations correctly.

Negative work occurs when the force has a component opposite to the displacement, which by the work-energy theorem decreases the object's kinetic energy. The statement that the force does no work is wrong because zero work requires the force to be exactly perpendicular to displacement, not opposite to it. Recognizing the sign of work tells you immediately whether an object is speeding up or slowing down.

Elastic potential energy equals \(\frac{1}{2}kx^2\) because it is the area under the linear force-versus-displacement graph described by Hooke's law, \(F = kx\). The choice \(PE_s = kx\) is wrong because that expression is simply the spring force itself, not the energy stored. Always remember spring energy depends on the square of displacement, so doubling compression quadruples stored energy.

Since \(KE = \frac{1}{2}mv^2\), kinetic energy depends on the square of speed, so doubling \(v\) multiplies \(KE\) by \(2^2 = 4\). The choice of a factor of 2 is wrong because that would only be true if kinetic energy depended linearly on speed, which it does not. This squared relationship is a frequent source of AP exam trick questions, so always square the velocity ratio.

A conservative force, such as gravity or an ideal spring force, does work that depends only on the endpoints of motion because it has an associated potential energy function. The choice describing path dependence is wrong because that describes nonconservative forces like friction, which dissipate energy differently depending on the path length. Recognizing conservative forces lets you use potential energy and conservation of mechanical energy to simplify problems.

Work is given by \(W = Fd\cos\theta\), and when the force is perpendicular to displacement \(\theta = 90^\circ\), making \(\cos\theta = 0\) and the work zero. The answer involving the object's mass is wrong because work depends on force, displacement, and angle, not directly on mass. This is why centripetal force does no work on an object moving in a circle at constant speed.

Power measures the rate of doing work, \(P = \frac{W}{t}\), and joules per second define the watt, the SI unit of power. The Joule is wrong because it measures energy or work itself, not the rate at which that energy is transferred. Distinguishing joules from watts is critical because many exam problems ask you to convert between energy and power using elapsed time.

Mechanical energy is defined as $E_{mech} = KE + PE$, representing the total energy associated with an object's motion and position in a conservative force field. The choice referencing only kinetic energy is wrong because it ignores stored potential energy, which is equally part of the mechanical energy total. In the absence of nonconservative forces like friction, total mechanical energy remains constant, a key idea for conservation of energy problems.

Work is the dot product of force and displacement vectors, which simplifies to \(W = Fd\cos\theta\) because only the component of force along the displacement direction contributes to work. The sine version is wrong because that formula is instead used to find torque or the perpendicular force component, not work. Always use cosine for work problems involving an angled force.

Kinetic friction is nonconservative because the work it does depends on the total path length traveled, converting mechanical energy irreversibly into thermal energy. Gravity is wrong to list here because it is a conservative force whose work depends only on the change in height, not the path taken. Recognizing nonconservative forces tells you that mechanical energy is not conserved and you must instead track energy lost to heat.

In uniform circular motion, the centripetal force is always directed perpendicular to the velocity, so \(W = Fd\cos(90^\circ) = 0\) at every instant, meaning no net work is done. The answer claiming positive work is wrong because the presence of a net force does not guarantee positive work unless it has a component along the displacement. This illustrates that constant speed implies constant kinetic energy, consistent with zero net work by the work-energy theorem.

Since the force and displacement are aligned, \(W = Fd\cos(0^\circ) = (20)(5)(1) = 100\text{ J}\). The choice of \(20\text{ J}\) is wrong because it only reflects the force magnitude and ignores the distance over which the force acts. Always multiply force by displacement, not just report one of the given quantities, when the angle between them is zero.

At maximum height all kinetic energy converts to potential energy, so $\frac{1}{2}mv^2 = mgh$ gives \(h = \frac{v^2}{2g} = \frac{400}{20} = 20\text{ m}\). The choice of \(10\text{ m}\) is wrong because it corresponds to forgetting to divide by the factor of two from the kinetic energy formula. This energy method avoids needing kinematics equations and directly ties speed to height through conservation of mechanical energy.

Using \(PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(150)(0.2)^2 = 3\text{ J}\), the stored energy is found by squaring the compression distance. The choice \(15\text{ J}\) is wrong because it results from forgetting to square the \(0.2\text{ m}\) compression before multiplying. Since \(x\) is squared, small changes in compression distance have an amplified effect on stored spring energy.

The work done equals the change in kinetic energy, \(\Delta KE = \frac{1}{2}(1000)(20)^2 = 200000\text{ J}\), and average power is \(P = \frac{W}{t} = \frac{200000}{5} = 40000\text{ W}\). The choice \(20000\text{ W}\) is wrong because it comes from mistakenly halving the correct kinetic energy value before dividing by time. This problem shows how power calculations often require first finding the work or energy change using the work-energy theorem.

The work-energy theorem gives \(W = Fd = (10)(4) = 40\text{ J} = \frac{1}{2}(2)v^2\), so \(v = \sqrt{40} \approx 6.3\text{ m/s}\). The choice \(4.5\text{ m/s}\) is wrong because it comes from incorrectly dividing the work by mass instead of solving for velocity through the kinetic energy formula. This problem shows the work-energy theorem's power to find speed directly from net work without needing acceleration or time.

The friction force is \(f = \mu mg = (0.3)(5)(10) = 15\text{ N}\), and since friction opposes motion, \(W_f = -fd = -(15)(4) = -60\text{ J}\). The choice \(-20\text{ J}\) is wrong because it fails to correctly multiply the friction force by the full distance traveled. Friction always removes mechanical energy from a system, converting it into thermal energy, which is why its work is always negative when opposing motion.

Gravitational potential energy change depends only on vertical height gained, $\Delta PE = mgh = (3)(10)(2) = 60\text{ J}$, regardless of the incline's length or angle. The choice \(30\text{ J}\) is wrong because it results from forgetting to include the full mass value in the calculation. Since gravity is conservative, only the vertical height change matters, not the path length along the incline.

Conservation of energy gives $mgh = \frac{1}{2}mv^2$, so \(v = \sqrt{2gh} = \sqrt{2(10)(3)} = \sqrt{60} \approx 7.75\text{ m/s}\). The choice \(6\text{ m/s}\) is wrong because it results from an arithmetic error in evaluating the square root of \(60\). This result confirms that on a frictionless incline, final speed depends only on the height dropped, not the incline's angle or length.

At constant velocity the applied force equals the weight, so the work done equals the gravitational potential energy gained, $W = mgh = (10)(10)(2) = 200\text{ J}$. The choice \(100\text{ J}\) is wrong because it omits a factor of the gravitational acceleration in the calculation. When lifting at constant speed, the work by the applied force always equals the change in gravitational potential energy.

Since the tension in the string does no work, mechanical energy is conserved, giving \(v = \sqrt{2gh} = \sqrt{2(10)(0.8)} = \sqrt{16} = 4\text{ m/s}\). The choice \(2.83\text{ m/s}\) is wrong because it corresponds to an incorrect height value being substituted into the square root. Pendulum problems are a classic application of energy conservation because the constraint force does no work throughout the swing.

Gravity acts vertically while the block's displacement is entirely horizontal, so the angle between the two vectors is \(90^\circ\), making \(W = Fd\cos(90^\circ) = 0\text{ J}\). The choice \(200\text{ J}\) is wrong because it mistakenly treats gravity as contributing work whenever motion occurs, without checking the direction of displacement relative to the force. A force does zero work whenever it is entirely perpendicular to an object's path, regardless of the force's magnitude.

The change in kinetic energy is \(\Delta KE = \frac{1}{2}(1200)(20^2 - 10^2) = 180000\text{ J}\), and dividing by time gives \(P = \frac{180000}{6} = 30000\text{ W}\). The choice \(60000\text{ W}\) is wrong because it results from doubling the correct answer by mistakenly using the final speed squared alone instead of the difference of squares. This problem shows that average power over an interval requires the net change in kinetic energy, not just the final kinetic energy.

Since \(KE \propto v^2\), doubling the speed while keeping mass constant multiplies kinetic energy by \(2^2 = 4\), giving a ratio of \(4\). The choice of \(2\) is wrong because it treats kinetic energy as directly proportional to speed rather than to speed squared. Always square the velocity ratio when comparing kinetic energies of objects with identical mass.

The ball has fallen \(20 - 12 = 8\text{ m}\), so the fraction of total energy converted to kinetic energy is \(\frac{8}{20} = 40\%\), since total mechanical energy stays constant during free fall. The choice \(60\%\) is wrong because it corresponds to the fraction of potential energy remaining rather than the fraction converted to kinetic energy. This problem illustrates that as an object falls, potential energy steadily converts into kinetic energy while their sum stays fixed.

All spring potential energy converts to kinetic energy on a frictionless surface, so \(\frac{1}{2}(100)(0.3)^2 = \frac{1}{2}(0.5)v^2\) gives \(v = \sqrt{18} \approx 4.24\text{ m/s}\). The choice \(3\text{ m/s}\) is wrong because it results from an arithmetic slip when solving for \(v\) from the energy equation. Spring launch problems always rely on setting elastic potential energy equal to kinetic energy when friction is absent.

At constant velocity, the crane's force equals the object's weight, so power is $P = Fv = mgv = (200)(10)(0.5) = 1000\text{ W}$. The choice \(500\text{ W}\) is wrong because it comes from omitting the gravitational acceleration factor in computing the required lifting force. For constant-velocity lifting, power always equals weight multiplied by the constant speed.

The friction force is \(f = \mu mg = (0.2)(4)(10) = 8\text{ N}\), and setting the initial kinetic energy equal to the work done by friction gives \(\frac{1}{2}(4)(36) = 8d\), so \(d = 9\text{ m}\). The choice \(4.5\text{ m}\) is wrong because it results from forgetting the factor of one-half in the kinetic energy expression before dividing by the friction force. This problem demonstrates using the work-energy theorem to find stopping distance without needing time or acceleration explicitly.

The relevant height change is \(45 - 20 = 25\text{ m}\), so conservation of energy gives \(v = \sqrt{2g\Delta h} = \sqrt{2(10)(25)} = \sqrt{500} \approx 22.4\text{ m/s}\). The choice \(15.8\text{ m/s}\) is wrong because it results from using the wrong height difference in the square root calculation. On frictionless tracks, only the net height drop matters for determining speed, not the shape of the track between those heights.

The net work equals the sum of the work done by each force, \((50)(10) + (-20)(10) = 500 - 200 = 300\text{ J}\), which by the work-energy theorem equals the box's change in kinetic energy. The choice \(500\text{ J}\) is wrong because it only accounts for the applied force's work and ignores the negative work done by friction. Always sum the work done by every force acting on an object to find the net work, not just the applied force alone.

As the ball rises, gravity does negative work on it, decreasing its kinetic energy while simultaneously increasing its gravitational potential energy, and because gravity is conservative, the total mechanical energy \(KE + PE\) remains constant. The choice stating both increase together is wrong because energy is being transferred from kinetic to potential form, not created in both forms simultaneously. This trade-off between kinetic and potential energy, with a constant total, is the essence of mechanical energy conservation.

All gravitational potential energy converts to spring potential energy at maximum compression, so $mgh = \frac{1}{2}kx^2$ gives \((4)(10)(2) = \frac{1}{2}(500)x^2\), and solving yields \(x = \sqrt{0.32} \approx 0.566\text{ m}\). The choice \(0.4\text{ m}\) is wrong because it results from an arithmetic error when solving the quadratic equation for \(x\). This problem combines two energy storage mechanisms, gravitational and elastic, both governed by the same overarching conservation of energy principle.

At the highest point, only the vertical velocity component becomes zero while the horizontal velocity component remains constant throughout the flight, so kinetic energy reaches a minimum but nonzero value determined by that horizontal speed. The choice claiming kinetic energy is exactly zero is wrong because it incorrectly assumes the object momentarily stops entirely, which only happens for purely vertical projectile motion. This distinction between vertical and horizontal velocity components is essential for correctly analyzing energy in two-dimensional projectile motion.

The net gravitational potential energy lost equals the kinetic energy gained by both masses, \((m_2 - m_1)gd = \frac{1}{2}(m_1+m_2)v^2\), giving \((2)(10)(2) = \frac{1}{2}(8)v^2\), so \(v = \sqrt{10} \approx 3.16\text{ m/s}\). The choice \(2.5\text{ m/s}\) is wrong because it fails to include both masses' kinetic energy in the total energy balance. This problem shows that energy methods can bypass tension forces entirely, which is often faster than a full force analysis for connected systems.

The kinetic energy at the bottom of the incline equals $mgh = (2)(10)(5) = 100\text{ J}$, and this is dissipated entirely by friction, \(f d = \mu mg d = (0.25)(2)(10)d = 5d\), so setting \(5d = 100\) gives \(d = 20\text{ m}\). The choice \(10\text{ m}\) is wrong because it results from using the wrong friction force value in the energy dissipation equation. This two-stage problem shows how energy gained on a frictionless section can be tracked into a frictional section using a single unified energy balance.

The work done by a variable force equals the area under the force-versus-position graph, which here is a triangle with area \(\frac{1}{2}(5)(40) = 100\text{ J}\). The choice \(200\text{ J}\) is wrong because it fails to apply the factor of one-half needed for a triangular area rather than a rectangular one. Whenever force varies with position, computing work requires finding the area under the curve rather than a simple multiplication.

The height dropped is \(h = d\sin(30^\circ) = 2\text{ m}\), giving gravitational energy $mgh = 100\text{ J}$, and since the final kinetic energy is only \(\frac{1}{2}(5)(16) = 40\text{ J}\), friction dissipated \(60\text{ J}\) over the incline distance, so solving \(60 = \mu mg\cos(30^\circ)(4)\) gives \(\mu \approx 0.35\). The choice \(0.50\) is wrong because it results from neglecting the \(\cos(30^\circ)\) factor that accounts for the normal force on an incline. Finding friction coefficients from energy loss requires carefully separating the normal force component from the total weight on an inclined surface.

The car's kinetic energy, \(\frac{1}{2}(1000)(20)^2 = 200000\text{ J}\), is entirely absorbed by the deformation, so using \(W = Fd\) gives \(F = \frac{200000}{0.5} = 400000\text{ N} = 400\text{ kN}\). The choice \(200\text{ kN}\) is wrong because it corresponds to forgetting to divide by the relatively small deformation distance, which greatly amplifies the resulting force. This calculation shows why crumple zones are engineered to increase stopping distance, since a larger \(d\) reduces the average force for the same energy absorbed.

The spring's stored energy is \(\frac{1}{2}(1000)(0.2)^2 = 20\text{ J}\), and setting this equal to gravitational potential energy at maximum height gives \(h = \frac{20}{(0.5)(10)} = 4\text{ m}\). The choice \(2\text{ m}\) is wrong because it results from doubling the mass value incorrectly when solving for height. This problem demonstrates chaining two different potential energy forms through a single conserved total energy value.

Since $dKE = F\,dx$ from the work-energy theorem, the slope of a kinetic energy versus position graph, \(\frac{d(KE)}{dx}\), directly equals the net force acting on the object. The choice of velocity is wrong because velocity cannot be extracted directly from a slope of this particular graph without first solving for kinetic energy at a point and applying the mass. Interpreting slopes of energy-versus-position graphs as forces is a valuable graphical skill for multi-representation AP Physics questions.

Gravity is a conservative force, meaning the work it does depends only on the change in vertical height between the start and end points, not on the specific path taken, so the work is identical for both paths. The choice claiming the curved path involves more work is wrong because path length alone does not determine work for a conservative force. This path-independence property is what allows gravitational potential energy to be defined as a well-behaved function of position alone.

At the lowest point all gravitational potential energy lost, $mgh = (60)(10)(30) = 18000\text{ J}$, has converted into elastic potential energy in the stretched cord, so setting \(\frac{1}{2}kx^2 = 18000\) with \(x = 10\text{ m}\) gives \(k = \frac{2(18000)}{100} = 360\text{ N/m}\). The choice \(180\text{ N/m}\) is wrong because it results from forgetting the factor of two when solving the elastic potential energy equation for \(k\). This problem shows how a full fall height, not just the stretch distance, must be used since gravitational energy is lost over the entire \(30\text{ m}\) descent.

At the height of \(10\text{ m}\) the frictionless descent has converted energy so that \(KE = mg\Delta h = (400)(10)(20) = 80000\text{ J}\), and after losing \(20\%\) of this to friction, the remaining kinetic energy is \(64000\text{ J} = \frac{1}{2}(400)v^2\), giving \(v = \sqrt{320} \approx 17.9\text{ m/s}\). The choice \(16\text{ m/s}\) is wrong because it results from incorrectly removing \(20\%\) of the speed rather than \(20\%\) of the kinetic energy before recomputing velocity. This problem requires combining conservation of energy on the frictionless section with a direct percentage energy loss on the rough section, a common multi-step AP exam structure.