Understand Work, Energy & Power - IB Physics

1. An object starts from rest. The net work done on the object is $W$. After this work is done, the object has momentum $p$. What is the net work required to be done on the object, starting from rest, to give it a momentum of $2p$?
   Answer: $4W$ By the work-energy theorem, the net work done on an object equals its change in kinetic energy, $W = \Delta E_k$. Since the object starts from rest, $W = E_k$. The kinetic energy can be expressed in terms of momentum $p$ and mass $m$ as $E_k = p^2 / (2m)$. So, $W = p^2 / (2m)$. We want to find the new work $W'$ required to achieve a momentum of $p' = 2p$. The new kinetic energy will be $E'_k = (p')^2 / (2m) = (2p)^2 / (2m) = 4p^2 / (2m) = 4(p^2 / (2m))$. Since $W = p^2 / (2m)$, the new work is $W' = 4W$.
2. A spring with spring constant $k$ is compressed by distance $x_0$ and then used to launch a projectile of mass $m$ at angle $45°$ to the horizontal. What is the maximum height reached by the projectile?
   Answer: $\frac{kx_0^2}{4mg}$ This problem combines elastic potential energy, projectile motion, and energy conservation. When you see a spring launching a projectile, think about the energy transformations: elastic potential energy converts to kinetic energy, then kinetic energy converts to gravitational potential energy at maximum height. The compressed spring stores elastic potential energy $E_p = \frac{1}{2}kx_0^2$. This energy becomes the projectile's initial kinetic energy: $\frac{1}{2}mv_0^2 = \frac{1}{2}kx_0^2$, so $v_0 = \sqrt{\frac{k}{m}}x_0$. For projectile motion at 45°, the initial vertical velocity component is $v_{0y} = v_0 \sin(45°) = \frac{v_0}{\sqrt{2}} = \frac{x_0}{\sqrt{2}}\sqrt{\frac{k}{m}}$. At maximum height, all vertical kinetic energy converts to gravitational potential energy: $\frac{1}{2}mv_{0y}^2 = mgh_{max}$. Substituting: $h_{max} = \frac{v_{0y}^2}{2g} = \frac{1}{2g} \cdot \frac{x_0^2}{2} \cdot \frac{k}{m} = \frac{kx_0^2}{4mg}$. Answer B is correct. Answer A ($\frac{kx_0^2}{2mg}$) incorrectly uses the full initial velocity instead of just the vertical component. Answer C ($\frac{kx_0^2}{mg}$) forgets both the $\frac{1}{2}$ factor from energy conversion and the vertical component reduction. Answer D ($\frac{kx_0^2}{8mg}$) includes an extra factor of 2 error, possibly from incorrectly applying the sine function. Remember: projectile motion problems launched at 45° always involve the factor $\frac{1}{\sqrt{2}}$ for velocity components, which becomes $\frac{1}{2}$ when squared in energy calculations.
3. A water pump lifts water at a rate of 0.50 m³ per minute from a well. The water is lifted 10 m and ejected with a speed of 4.0 m s⁻¹. What is the minimum useful power output of the pump? The density of water is 1000 kg m⁻³. Use $g = 10\ \text{m s}^{-2}$.
   Answer: 900 W The pump must provide power to increase both the gravitational potential energy and the kinetic energy of the water. First, find the mass flow rate in kg s⁻¹: $\dot{m} = \rho \times \text{volume rate} = 1000\ \text{kg m}^{-3} \times (0.50\ \text{m}^3 / 60\ \text{s}) = 25/3\ \text{kg s}^{-1}$. The power to increase GPE is $P_{GPE} = \dot{m}gh = (25/3)(10)(10) = 2500/3\ \text{W}$. The power to increase KE is $P_{KE} = \frac{1}{2}\dot{m}v^2 = \frac{1}{2}(25/3)(4.0)^2 = \frac{1}{2}(25/3)(16) = 200/3\ \text{W}$. The total minimum power output is the sum: $P_{total} = P_{GPE} + P_{KE} = 2500/3 + 200/3 = 2700/3 = 900\ \text{W}$.
4. An electric motor that is 75% efficient has a useful mechanical power output of 300 W. At what rate is heat dissipated by the motor?
   Answer: 100 W Efficiency $\eta = P_{out} / P_{in}$. The electrical power input is $P_{in} = P_{out} / \eta = 300\ \text{W} / 0.75 = 400\ \text{W}$. The rate of heat dissipation is the wasted power, which is the difference between the power input and the useful power output: $P_{wasted} = P_{in} - P_{out} = 400\ \text{W} - 300\ \text{W} = 100\ \text{W}$. This wasted power is primarily converted into thermal energy (heat).
5. A car of mass $m$ starts from rest on a horizontal road and is accelerated by an engine that delivers constant power $P$. Air resistance is negligible. Which statement best describes the acceleration of the car?
   Answer: The acceleration is initially large and decreases as speed increases. Power is given by $P = Fv$, where $F$ is the driving force and $v$ is the instantaneous speed. From Newton's second law, $F = ma$. Substituting this into the power equation gives $P = (ma)v$. Rearranging for acceleration $a$ gives $a = P / (mv)$. Since $P$ and $m$ are constant, the acceleration $a$ is inversely proportional to the speed $v$. As the car starts from rest and its speed increases, its acceleration must decrease.
6. The engine is modified to increase its efficiency. If the energy input remains the same, how will the Sankey diagram change?
   Answer: The useful work arrow will be wider and the waste heat arrow will be narrower. Efficiency is the ratio of useful work output to total energy input. If the efficiency increases with the same energy input, a larger fraction of the input energy must be converted into useful work. In a Sankey diagram, the width of the arrow is proportional to the amount of energy. Therefore, the arrow for useful work must become wider. By the principle of conservation of energy, if the useful work output increases, the waste heat must decrease (since Input = Useful Work + Waste Heat). Thus, the waste heat arrow will become narrower.
7. A motor lifts a $500\text{ kg}$ load at constant velocity through a height of $20\text{ m}$ in $40\text{ s}$. If the motor operates at $75\%$ efficiency, what is the electrical power input to the motor?
   Answer: $3.27\text{ kW}$ Mechanical power output = $\frac{mgh}{t} = \frac{500 \times 9.8 \times 20}{40} = 2450\text{ W}$. Since efficiency = $\frac{P_{out}}{P_{in}}$, electrical power input = $\frac{2450}{0.75} = 3267\text{ W} = 3.27\text{ kW}$. Choice B gives the mechanical power output instead of input. Choice C incorrectly multiplies by efficiency. Choice D uses an incorrect calculation.
8. A projectile of mass $m$ is launched with an initial kinetic energy $E$. At the highest point of its trajectory, its kinetic energy is $E/4$. Air resistance is negligible. What was the launch angle with respect to the horizontal?
   Answer: 60° Let the launch velocity be $v$ at an angle $\theta$. The initial kinetic energy is $E = \frac{1}{2}mv^2$. The velocity has components $v_x = v\cos\theta$ and $v_y = v\sin\theta$. At the highest point of the trajectory, the vertical component of velocity is zero, but the horizontal component $v_x$ remains unchanged. The kinetic energy at the highest point is $E_{top} = \frac{1}{2}m v_x^2 = \frac{1}{2}m(v\cos\theta)^2 = (\frac{1}{2}mv^2)\cos^2\theta = E\cos^2\theta$. We are given that $E_{top} = E/4$. Therefore, $E\cos^2\theta = E/4$, which simplifies to $\cos^2\theta = 1/4$. Taking the square root gives $\cos\theta = 1/2$. The angle $\theta$ is 60°.
9. A 2.0 kg mass is moved from point X at a height of 1.0 m above the ground to point Y at a height of 3.0 m. Path 1 is a direct vertical lift. Path 2 is a 10 m long frictionless ramp. Let $W_1$ be the work done against gravity for Path 1 and $W_2$ be the work done against gravity for Path 2. Which statement is correct?
   Answer: $W_1 = W_2$ The gravitational force is a conservative force. This means the work done against gravity depends only on the change in vertical height (potential energy), not on the path taken. The change in height is $\Delta h = 3.0\ \text{m} - 1.0\ \text{m} = 2.0\ \text{m}$ for both paths. Therefore, the work done against gravity is the same for both paths: $W = mg\Delta h$.
10. A 0.50 kg block is placed against a horizontal spring of constant $k = 800\ \text{N m}^{-1}$ which has been compressed by 0.20 m. The spring is released, and the block slides across a frictionless horizontal surface and then up a frictionless ramp. What is the maximum vertical height $h$ the block reaches on the ramp? Use $g=10\ \text{m s}^{-2}$.
    Answer: 3.2 m This problem uses the conservation of energy. The initial energy is stored as elastic potential energy in the spring. This is converted into kinetic energy and then into gravitational potential energy. The initial elastic potential energy is $E_H = \frac{1}{2}k\Delta x^2 = \frac{1}{2}(800)(0.20)^2 = 400 \times 0.04 = 16\ \text{J}$. At the maximum height on the ramp, all of this energy will have been converted to gravitational potential energy, $E_p = mgh$. So, $mgh = 16\ \text{J}$. $h = 16 / (mg) = 16 / (0.50 \times 10) = 16 / 5 = 3.2\ \text{m}$.
11. A fossil fuel power plant has three stages of energy conversion. The boiler has an efficiency of 80%, the turbine has an efficiency of 50%, and the generator has an efficiency of 95%. What is the overall efficiency of the plant in converting the chemical energy of the fuel into electrical energy?
    Answer: 38% The overall efficiency of a multi-stage process is the product of the efficiencies of the individual stages. The efficiencies must be expressed as decimals for multiplication. Overall efficiency $\eta_{total} = \eta_1 \times \eta_2 \times \eta_3 = 0.80 \times 0.50 \times 0.95 = 0.38$. Expressed as a percentage, this is 38%.
12. A block is pulled up a $30°$ incline by a force $F$ parallel to the incline. The coefficient of kinetic friction is $0.2$. If the block moves at constant velocity, what fraction of the work done by force $F$ is dissipated as heat due to friction?
    Answer: $0.26$ When analyzing work and energy on inclined planes with friction, you need to identify where the applied force's energy goes. Since the block moves at constant velocity, the net force is zero, meaning the applied force $F$ exactly balances both the gravitational component down the incline and friction. Let's set up the force balance. The weight component down the incline is $mg\sin(30°) = 0.5mg$. The friction force is $\mu_k mg\cos(30°) = 0.2mg \times \frac{\sqrt{3}}{2} = 0.1\sqrt{3}mg \approx 0.173mg$. Therefore, $F = mg\sin(30°) + \mu_k mg\cos(30°) = 0.5mg + 0.173mg = 0.673mg$. When the block moves distance $d$ up the incline, the work done by force $F$ is $W_F = Fd = 0.673mgd$. The work dissipated by friction is $W_{friction} = \mu_k mg\cos(30°) \times d = 0.173mgd$. The fraction dissipated as heat is $\frac{W_{friction}}{W_F} = \frac{0.173mgd}{0.673mgd} = \frac{0.173}{0.673} = 0.26$. Answer B (0.26) is correct. Answer A (0.50) incorrectly assumes friction equals the gravitational component. Answer C (0.20) mistakenly uses just the coefficient of friction without considering the geometry. Answer D (0.65) appears to invert the calculation or use an incorrect force relationship. Remember: on inclined planes with friction, always resolve forces parallel and perpendicular to the surface, and account for both gravitational and frictional components when calculating work distribution.
13. A person carries a 20 kg box horizontally for 10 m at a constant velocity. What is the work done by the person on the box?
    Answer: 0 J Work is done when a force causes displacement in the direction of the force. The person exerts an upward vertical force on the box to counteract gravity. The displacement of the box is horizontal. Since the force is perpendicular to the displacement, the angle $\theta$ between them is 90°. The work done is $W = Fd\cos\theta = Fd\cos(90^\circ) = 0\ \text{J}$.
14. A pendulum bob of mass $m$ swings from rest at height $h$ above its lowest point. When the string makes an angle $\theta$ with the vertical, what is the tension in the string?
    Answer: $mg(3\cos\theta - 2\cos\theta_0)$ This question tests your understanding of circular motion dynamics combined with energy conservation for a pendulum. When analyzing pendulum motion, you need to consider both the gravitational forces and the centripetal force requirements. First, establish the initial conditions. The pendulum starts from rest at angle $\theta_0$ where $\cos\theta_0 = 1 - \frac{h}{L}$ (L is the string length). Using energy conservation, when the string makes angle $\theta$, the speed is $v^2 = 2gL(\cos\theta - \cos\theta_0)$. At any position, two forces act on the bob: weight $mg$ downward and tension $T$ along the string. The radial component equation gives us: $T - mg\cos\theta = \frac{mv^2}{L}$. Substituting the velocity from energy conservation: $T = mg\cos\theta + \frac{m \cdot 2gL(\cos\theta - \cos\theta_0)}{L} = mg\cos\theta + 2mg(\cos\theta - \cos\theta_0) = mg(3\cos\theta - 2\cos\theta_0)$. Option A, $mg(2\cos\theta - \cos\theta_0)$, incorrectly omits the gravitational component along the string direction. Option B, $mg\cos\theta$, only accounts for the gravitational component but ignores the centripetal force requirement entirely. Option D, $mg(\cos\theta + 2\cos\theta_0)$, has the wrong sign for the initial position term and would give unphysical results. For pendulum problems, always remember to apply both energy conservation (to find speed) and Newton's second law in the radial direction (to find tension). The tension must provide both the radial component of gravitational support and the centripetal force.
15. A block slides down a frictionless inclined plane of height $h$ and then compresses a spring at the bottom. If the spring constant is $k$ and the block comes to rest after compressing the spring by distance $x$, what is the mass of the block?
    Answer: $m = \frac{kx^2}{2gh}$ Using conservation of energy: Initial gravitational potential energy equals final elastic potential energy. $mgh = \frac{1}{2}kx^2$. Solving for mass: $m = \frac{kx^2}{2gh}$. Choice B inverts the relationship incorrectly. Choice C omits the square on $x$. Choice D is missing the factor of 2.
16. A power station produces the same amount of useful electrical energy per day by burning either 10 tonnes of fuel A (energy density 45 MJ kg⁻¹) or 12 tonnes of fuel B (energy density 30 MJ kg⁻¹). What is the ratio of the efficiency when using Fuel A to the efficiency when using Fuel B ($\eta_A / \eta_B$)?
    Answer: 0.80 Efficiency is $\eta = E_{out} / E_{in}$. Since $E_{out}$ is the same for both fuels, the ratio of efficiencies is $\eta_A / \eta_B = (E_{out}/E_{in,A}) / (E_{out}/E_{in,B}) = E_{in,B} / E_{in,A}$. First, calculate the total energy input for each fuel. $E_{in,A} = (10 \times 1000\ \text{kg}) \times (45\ \text{MJ kg}^{-1}) = 450000\ \text{MJ}$. $E_{in,B} = (12 \times 1000\ \text{kg}) \times (30\ \text{MJ kg}^{-1}) = 360000\ \text{MJ}$. The ratio is $\eta_A / \eta_B = 360000 / 450000 = 36/45 = 4/5 = 0.80$.
17. Particle X has mass $m$ and particle Y has mass (4m). Both particles have the same non-zero kinetic energy. What is the ratio of the magnitude of the momentum of Y to the magnitude of the momentum of X ($p_Y/p_X$)?
    Answer: 2 The relationship between kinetic energy $E_k$ and momentum $p$ is $E_k = p^2 / (2m)$. Rearranging for momentum gives $p = \sqrt{2mE_k}$. Since both particles have the same kinetic energy $E_k$, the momentum is proportional to the square root of the mass ($p \propto \sqrt{m}$). The ratio of the momenta is therefore $p_Y/p_X = \sqrt{m_Y} / \sqrt{m_X} = \sqrt{4m} / \sqrt{m} = \sqrt{4} = 2$.
18. A vehicle travels at a constant speed $v$ against a total resistive force $F$. The engine provides power $P$. The resistive force is proportional to the square of the speed ($F \propto v^2$). What is the new power $P'$ required to travel at a constant speed of $2v$?
    Answer: 8$P$ The power required to overcome a force $F$ at speed $v$ is $P = Fv$. We are given that the resistive force $F$ is proportional to $v^2$, so we can write $F = kv^2$ for some constant $k$. Substituting this into the power equation gives $P = (kv^2)v = kv^3$. Therefore, power is proportional to the cube of the speed. If the speed is doubled from $v$ to $2v$, the new power $P'$ will be $k(2v)^3 = k(8v^3) = 8(kv^3) = 8P$.
19. An external force compresses a spring that obeys Hooke's Law by 0.20 m. The force required to hold the spring at this compression is 50 N. How much work was done by the external force to compress the spring?
    Answer: 5.0 J The work done in compressing a spring is equal to the elastic potential energy stored in it, $E_H = \frac{1}{2}k\Delta x^2$. First, find the spring constant $k$ from Hooke's Law, $F = k\Delta x$. So, $k = F/\Delta x = 50\ \text{N} / 0.20\ \text{m} = 250\ \text{N m}^{-1}$. Now, calculate the work done: $W = \frac{1}{2}(250\ \text{N m}^{-1})(0.20\ \text{m})^2 = \frac{1}{2}(250)(0.04) = 5.0\ \text{J}$. A common mistake is to calculate work as $W = F\Delta x = 50 \times 0.20 = 10\ \text{J}$, which is incorrect because the force is not constant during the compression.
20. A box is pushed by a person around a square horizontal path of side length 5 m, returning to its starting point. A constant kinetic friction force of 10 N opposes the motion. What is the net work done by the friction force on the box?
    Answer: –200 J Work done by friction is given by $W_f = -F_f \times d$, where $d$ is the total distance travelled. The total distance is the perimeter of the square, which is $4 \times 5\ \text{m} = 20\ \text{m}$. The work done by friction is therefore $W_f = -10\ \text{N} \times 20\ \text{m} = -200\ \text{J}$. The negative sign indicates that the force opposes the displacement.

Start this quiz