Work and Energy — Free Physics Review Games.

Work equals force multiplied by displacement in the direction of the force: W = Fd.

The Joule (J) is the SI unit of both energy and work.

Kinetic energy is the energy an object possesses due to its motion: KE = 1/2 mv^2.

Gravitational potential energy is stored energy based on an object's position: PE = mgh.

The law of conservation of energy states that total energy in a closed system remains constant; it can only change forms.

W = Fd = 20 N x 5 m = 100 J.

Power is the rate of doing work or transferring energy: P = W/t, measured in watts (W).

KE = 1/2 mv^2 = 1/2(2)(3^2) = 1/2(2)(9) = 9 J.

Elastic potential energy is stored in objects that are stretched or compressed, like springs: PE = 1/2 kx^2.

No work is done when the force is perpendicular to the displacement, such as carrying a box horizontally (gravity is vertical).

At the top while momentarily at rest, all energy is gravitational potential energy (PE = mgh) with zero kinetic energy.

Work = mgh = 200(9.8)(10) = 19,600 J. Time = W/P = 19600/1000 = 19.6 s.

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = delta KE.

As the pendulum swings down, gravitational potential energy converts to kinetic energy, reaching maximum KE at the lowest point.

Friction and other non-conservative forces always convert some useful energy into thermal energy, preventing any real machine from achieving 100% efficiency.

Work is defined as \(W = Fd\cos\theta\), which produces a single signed number with no direction, making it a scalar quantity. Force is wrong because force has both magnitude and direction and is represented as a vector in Newton's second law. Students should remember that energy and work are scalars, while quantities like force, velocity, and acceleration are vectors.

Work is given by \(W = Fd\cos\theta\), and when the force is perpendicular to displacement, \(\theta = 90^\circ\) so \(\cos\theta = 0\), giving zero work. The choice 'Maximum work' is wrong because maximum work occurs when the force is parallel to displacement (\(\theta = 0^\circ\)), not perpendicular. This is why centripetal forces do no work on orbiting objects, since they always act perpendicular to velocity.

Kinetic energy depends on the square of speed, \(KE = \frac{1}{2}mv^2\), so doubling \(v\) increases \(KE\) by a factor of \(2^2 = 4\). The choice 'It doubles' incorrectly assumes a linear relationship between speed and kinetic energy rather than a quadratic one. Always remember that kinetic energy scales with velocity squared, which has major implications for stopping distances and collision severity.

Gravitational potential energy near Earth's surface is defined as $PE = mgh$, where \(m\) is mass, \(g\) is gravitational acceleration, and \(h\) is height above a reference point. The formula \(\frac{1}{2}mv^2\) is wrong here because that expression defines kinetic energy, not potential energy. Students should keep gravitational PE, kinetic energy, and elastic PE formulas distinct since each depends on different variables.

In an isolated system without non-conservative forces like friction, total mechanical energy (kinetic plus potential) is conserved and stays constant, per the law of conservation of energy. The choice 'It continuously decreases' is wrong because that only happens when energy is lost to friction or other dissipative forces, which are excluded here. This principle lets you set initial mechanical energy equal to final mechanical energy to solve many physics problems.

Elastic potential energy depends on the square of the compression or stretch distance \(x\) in the formula \(PE = \frac{1}{2}kx^2\), so increasing \(x\) directly and significantly increases stored energy. 'The spring's natural length' is wrong because natural length does not appear in the elastic PE formula at all. Remember that both the spring constant \(k\) and displacement \(x\) determine elastic potential energy, with displacement having a squared effect.

Negative work occurs when the force component is opposite to displacement, since \(W = Fd\cos\theta\) becomes negative when \(\theta\) is between \(90^\circ\) and \(180^\circ\). 'A force acts perpendicular to displacement' is wrong because that condition gives zero work, not negative work. A common example is friction doing negative work on a sliding object, removing kinetic energy from the system.

Power is the rate of energy transfer, defined as \(P = \frac{W}{t}\), and its SI unit is the watt, equal to one joule per second. 'Joule' is wrong because that unit measures energy or work itself, not the rate at which it is transferred. Keep in mind that watts measure how quickly work is done, which is different from the total amount of work or energy involved.

Kinetic energy is specifically the energy an object has because it is moving, calculated using \(KE = \frac{1}{2}mv^2\). 'Energy stored due to an object's position' is wrong because that describes potential energy, not kinetic energy. Distinguishing energy due to motion versus energy due to position is essential for solving conservation of energy problems.

Since kinetic energy depends on velocity squared as \(KE = \frac{1}{2}mv^2\), and the book's velocity is zero, its kinetic energy must be exactly zero. 'A negative amount' is wrong because kinetic energy can never be negative, since mass is always positive and velocity squared is always non-negative. This reinforces that kinetic energy is always zero or positive, never negative, regardless of the object's mass.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy, expressed as $W_{net} = \Delta KE$. The equation $W = mgh$ is wrong because that expression calculates gravitational potential energy, not work in terms of kinetic energy change. This theorem is a powerful shortcut for solving problems involving forces, distances, and resulting speed changes.

Work is calculated as \(W = Fd\cos\theta\), and since the force is horizontal and in the direction of motion, \(\theta = 0^\circ\), giving \(W = 15 \times 4 = 60\ \text{J}\). The choice \(15\ \text{J}\) is wrong because it only reflects the force value and ignores the distance traveled entirely. Always multiply force by displacement (with the correct angle) to find work, rather than reporting just one variable.

Gravitational potential energy is $PE = mgh = 3 \times 10 \times 10 = 300\ \text{J}$, using the given mass, gravitational acceleration, and height. The choice \(30\ \text{J}\) is wrong because it omits the height factor, calculating only \(mg\) without multiplying by \(h\). Always confirm all three variables, mass, gravity, and height, are included when computing gravitational potential energy.

By conservation of energy, the kinetic energy gained equals the potential energy lost: $KE = mgh = 5 \times 10 \times 4 = 200\ \text{J}$. The choice \(50\ \text{J}\) is wrong because it fails to correctly multiply all three variables mass, gravity, and height together. This shortcut of setting $KE = PE_{lost}$ works whenever no non-conservative forces like air resistance are acting on a falling object.

Power is calculated as \(P = \frac{W}{t} = \frac{50{,}000}{10} = 5000\ \text{W}\), dividing the total work done by the time taken. The choice \(500{,}000\ \text{W}\) is wrong because it multiplies the work and time instead of dividing them, giving an inflated result. Remember that power always represents a rate, so time must be divided into the energy or work value, not multiplied.

Elastic potential energy is \(PE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = \frac{1}{2}(200)(0.01) = 1\ \text{J}\). The choice \(20\ \text{J}\) is wrong because it forgets to square the compression distance before multiplying, treating \(x\) as if it were linear rather than squared. Always square the displacement first in the elastic PE formula since the relationship is quadratic, not linear.

Since \(KE \propto v^2\), the ratio of kinetic energies equals the square of the speed ratio: \(\frac{16000}{4000} = 4 = \left(\frac{v_f}{v_i}\right)^2\), so the speed ratio is \(\sqrt{4} = 2\). The choice \(4\) is wrong because it confuses the energy ratio itself with the speed ratio, forgetting to take the square root. This reinforces that kinetic energy grows with the square of speed, so energy ratios must be square-rooted to find speed ratios.

When a moving ball compresses a spring, its kinetic energy is transformed into elastic potential energy stored in the compressed spring, illustrating an energy conversion. 'A ball rolls at constant speed across a flat floor' is wrong because no energy conversion is occurring there, as the kinetic energy stays constant with no change in speed. Recognizing these conversions between energy forms is key to applying conservation of energy in multi-stage problems.

Since the box moves at constant speed, the worker's applied force equals gravity's force, so the work done equals the change in gravitational potential energy, $W = mgh = 10 \times 10 \times 2 = 200\ \text{J}$. The choice \(20\ \text{J}\) is wrong because it omits multiplying by the height, only accounting for the weight of the box. When lifting objects at constant velocity, the work done against gravity always equals $mgh$.

Since \(KE = \frac{1}{2}mv^2\) and both objects share the same speed, kinetic energy is directly proportional to mass, so doubling the mass doubles the kinetic energy. 'B has four times the kinetic energy of A' is wrong because that would apply if speed, not mass, had doubled, since only speed appears squared in the formula. Remember that mass and kinetic energy scale linearly together, while speed and kinetic energy scale quadratically.

Friction is a non-conservative force that converts kinetic energy into thermal energy through microscopic collisions between surface particles, causing the block to slow and stop. 'It is completely destroyed and disappears' is wrong because energy cannot be destroyed, only transformed, per the law of conservation of energy. Always account for energy 'lost' to friction as thermal energy rather than assuming it vanishes from the system.

Using the work-energy theorem, $W_{net} = \Delta KE = 0 - \frac{1}{2}(2)(4)^2 = -16\ \text{J}$, since the cart's final kinetic energy is zero. The choice \(16\ \text{J}\) is wrong because it ignores the negative sign that indicates energy is being removed from the cart, not added. Negative work values are essential for representing forces that decelerate or remove energy from a system, such as brakes or friction.

At the peak of vertical motion, the object's velocity momentarily reaches zero, making kinetic energy zero, while height is at its maximum, making potential energy maximum. 'Kinetic and potential energy are both at their average values' is wrong because at this specific instant the energies are at their extremes, not averages. This illustrates the constant trade-off between kinetic and potential energy that keeps total mechanical energy conserved throughout the motion.

Work is \(W = Fd\cos\theta = 10 \times 5 \times \cos(60^\circ) = 50 \times 0.5 = 25\ \text{J}\), since \(\cos(60^\circ) = 0.5\). The choice \(50\ \text{J}\) is wrong because it neglects the cosine factor entirely, treating the force as if it were fully aligned with the displacement. Always apply the cosine of the angle between force and displacement, since only the component of force along the motion contributes to work.

Using conservation of energy, $mgh = \frac{1}{2}mv^2$, the mass cancels and \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\ \text{m/s}\). The choice \(14.1\ \text{m/s}\) is wrong because it corresponds to omitting the factor of 2 inside the square root, using \(\sqrt{gh}\) instead of \(\sqrt{2gh}\). This frictionless hill scenario shows that speed at the bottom depends only on height, not mass, when energy is conserved.

By the work-energy theorem, \(W = \Delta KE = \frac{1}{2}mv^2\), so \(8 = \frac{1}{2}(1)v^2\), giving \(v^2 = 16\) and \(v = 4\ \text{m/s}\). The choice \(8\ \text{m/s}\) is wrong because it mistakenly treats work as directly equal to speed rather than solving through the squared kinetic energy relationship. Always solve for velocity by isolating \(v^2\) first and then taking the square root, since kinetic energy depends on velocity squared.

Work requires displacement, and since the object remains stationary, \(d = 0\), making \(W = Fd\cos\theta\) equal to zero regardless of the friction force's magnitude. 'Static friction always acts perpendicular to the surface' is wrong because static friction actually acts parallel to the surface, opposing relative sliding tendency, not perpendicular to it. This highlights that work depends fundamentally on displacement, so any force acting on a motionless object does zero work.

Solving \(KE = \frac{1}{2}mv^2\) for velocity gives \(v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 6}{0.5}} = \sqrt{24} \approx 4.9\ \text{m/s}\). The choice \(12\ \text{m/s}\) is wrong because it comes from forgetting to take the square root after solving for \(v^2\), leaving an unfinished calculation. Always complete the algebra by taking the square root when solving for velocity from a kinetic energy equation.

As the block descends the frictionless incline, its height decreases while its speed increases, meaning gravitational potential energy is being converted into kinetic energy, consistent with conservation of mechanical energy. 'Kinetic energy converts into gravitational potential energy' is wrong because that describes the reverse process, which would occur if the block were moving upward instead of downward. On any frictionless slope, height loss directly corresponds to a kinetic energy gain of equal magnitude.

Since work equals \(W = Fd\cos\theta\) and both carts experience the same force and distance, they receive identical amounts of work, and by the work-energy theorem this means identical gains in kinetic energy regardless of mass. 'The heavier cart gains more kinetic energy' is wrong because kinetic energy gained depends on the work done, not directly on mass, though the heavier cart will end up moving slower to have the same energy. This distinction matters because although kinetic energy gained is equal, the resulting speeds differ since \(KE = \frac{1}{2}mv^2\) depends on mass differently for each cart.

As the pendulum swings down, all of its initial gravitational potential energy converts into kinetic energy by the time it reaches the lowest point, where height above the reference is minimized and speed is maximized. 'At the highest point of release' is wrong because at that point the bob is momentarily at rest, giving zero kinetic energy and maximum potential energy instead. This trade-off between height and speed throughout the swing is a classic demonstration of mechanical energy conservation.

Since $PE = mgh$ is proportional to height, the potential energy remaining at \(8\ \text{m}\) is \(\frac{8}{20} \times 100{,}000 = 40{,}000\ \text{J}\), so the kinetic energy is the difference, \(100{,}000 - 40{,}000 = 60{,}000\ \text{J}\). The choice \(40{,}000\ \text{J}\) is wrong because that value represents the remaining potential energy at that height, not the kinetic energy gained. This proportional height method is a fast way to solve conservation of energy problems without needing to know the exact mass.

By energy conservation, the initial potential energy $mgh = 4 \times 10 \times 5 = 200\ \text{J}$ must equal the work done by friction, \(W_f = \mu mg d\), so \(200 = \mu(4)(10)(10)\), giving \(\mu = \frac{200}{400} = 0.5\). The choice \(0.2\) is wrong because it likely comes from an arithmetic slip, such as dividing the wrong pair of numbers when isolating \(\mu\). This problem shows how conservation of energy can be combined with the friction work formula to solve for an unknown coefficient of friction.

The work required is $W = mgh = 200 \times 10 \times 15 = 30{,}000\ \text{J}$, and since \(P = \frac{W}{t}\), the time is \(t = \frac{W}{P} = \frac{30{,}000}{1000} = 30\ \text{s}\). The choice \(15\ \text{s}\) is wrong because it corresponds to forgetting to include the gravitational acceleration factor when calculating total work, effectively halving the true energy requirement. Whenever combining power and energy problems, always calculate the total work first before solving for time using the power equation.

The elastic potential energy is \(\frac{1}{2}kx^2 = \frac{1}{2}(500)(0.2)^2 = 10\ \text{J}\), and setting this equal to $mgh$ gives \(10 = (0.5)(10)h\), so \(h = \frac{10}{5} = 2\ \text{m}\). The choice \(1\ \text{m}\) is wrong because it likely results from forgetting to square the compression distance, using \(x\) linearly instead of \(x^2\) in the elastic energy formula. Multi-step conversion problems like this require carefully tracking energy as it transforms from one form to another while keeping total energy constant.

The work-energy theorem, $W_{net} = \Delta KE$, holds because Newton's second law shows the net force determines the object's acceleration, and integrating force over distance for the net force yields the total kinetic energy change regardless of how many individual forces contribute. 'It states that only friction forces contribute to kinetic energy changes' is wrong because any net force, including applied forces, gravity, and normal forces, can contribute to changes in kinetic energy, not friction alone. This theorem is powerful precisely because it lets you sum the work of every individual force to predict the overall change in speed.

Since air resistance is negligible, only conservative gravitational forces act on the pendulum, so total mechanical energy stays constant while kinetic and potential energy continuously convert into each other during the swing. 'It becomes zero exactly at the lowest point' is wrong because at the lowest point the kinetic energy actually reaches its maximum value, keeping total mechanical energy the same as at the start, not zero. This constant total energy is the hallmark of an idealized conservative system, useful for predicting speed at any point in the swing.

Real machines always experience some friction, air resistance, or internal resistive losses that convert a portion of input energy into heat or sound rather than useful output work, making perfect efficiency impossible. 'Machines always violate the law of conservation of energy' is wrong because efficiency losses do not violate conservation of energy at all, since the energy is simply transformed into non-useful forms rather than destroyed. Recognizing that efficiency is always less than 100 percent in practice, due to unavoidable dissipative forces, is essential for solving real-world energy transfer problems.

The horizontal force does \(W = 20 \times 3 = 60\ \text{J}\) of work, while gravity does negative work equal to $-mgh = -(2)(10)(1) = -20\ \text{J}$, so the net work is \(60 - 20 = 40\ \text{J}\). The choice \(60\ \text{J}\) is wrong because it only accounts for the applied horizontal force's work, ignoring the negative work done by gravity as the block rises vertically. In incline problems, always account for both the applied force's work and the work done against gravity to correctly find the net work and resulting kinetic energy change.

The initial potential energy is $mgh_1 = 0.2 \times 10 \times 5 = 10\ \text{J}$, and the final potential energy at the peak of the bounce is $mgh_2 = 0.2 \times 10 \times 3 = 6\ \text{J}$, so the energy lost is \(10 - 6 = 4\ \text{J}\). The choice \(6\ \text{J}\) is wrong because that value represents the remaining energy after the bounce, not the amount lost during the collision with the ground. Collisions like bouncing balls typically lose mechanical energy to sound, heat, and deformation, which is why the rebound height is always less than the drop height in real situations.

The initial potential energy is $mgh = 10 \times 10 \times 5 = 500\ \text{J}$, and the friction work is \(W_f = \mu mg d = 0.625 \times 10 \times 10 \times 8 = 500\ \text{J}\), so the energy balances exactly, confirming the box stops precisely at \(8\ \text{m}\). The choice 'No, since friction work exceeds the available potential energy' is wrong because the calculated friction work exactly matches the initial potential energy rather than exceeding it. This kind of check, verifying that energy lost to friction equals the energy initially available, is a reliable way to confirm consistency in multi-stage energy problems.

Since mechanical energy is conserved for both balls and they start with the same kinetic and potential energy from the same height and speed, they must have the same total energy just before impact, resulting in equal final speeds despite different directions of initial motion. 'The ball thrown upward hits the ground faster' is wrong because although it travels higher first, it eventually returns to the same starting height with the same speed it was launched, then falls the same distance as the other ball with identical energy transformations. This result demonstrates that conservation of energy depends only on height and speed magnitudes, not on the specific direction of initial velocity.

By the work-energy theorem, \(W = \Delta KE = \frac{1}{2}mv^2\), so \(54 = \frac{1}{2}(3)v^2\), giving \(v^2 = 36\) and \(v = 6\ \text{m/s}\). The choice \(18\ \text{m/s}\) is wrong because it comes from forgetting to divide by mass properly or skipping the square root step in solving for velocity. Even when a force varies with position, the work-energy theorem still applies directly to the total work done, without needing to know the exact force function in detail.

The energy needed at the top includes both kinetic energy (\(8000\ \text{J}\)) and potential energy at the loop's height ($mgh = 400 \times 10 \times 10 = 40{,}000\ \text{J}$), totaling \(48{,}000\ \text{J}\), which must equal the potential energy at the start, $mgH = 400 \times 10 \times H$, giving \(H = \frac{48{,}000}{4000} = 12\ \text{m}\). The choice \(10\ \text{m}\) is wrong because it only accounts for the loop's height and neglects the additional kinetic energy required to maintain contact with the track at the top. Loop-the-loop problems require combining both potential energy at height and the minimum kinetic energy needed for circular motion into a single conservation of energy equation.

The elastic potential energy is \(\frac{1}{2}kx^2 = \frac{1}{2}(250)(0.12)^2 = \frac{1}{2}(250)(0.0144) = 1.8\ \text{J}\), and setting this equal to kinetic energy gives \(1.8 = \frac{1}{2}(0.1)v^2\), so \(v^2 = 36\) and \(v = 6\ \text{m/s}\). The choice \(3\ \text{m/s}\) is wrong because it likely results from a square root error, such as taking the square root of half the correct \(v^2\) value instead of the full value. This spring-launch scenario is a classic application of setting elastic potential energy equal to kinetic energy under the assumption of no energy losses.