Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 60 km/h sees an obstacle. Her reaction time is 0.5 s, after which the car brakes with a constant deceleration of 5.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 6 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 20 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 95 N pulls the front 1 kg block; the 6 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is kicked from level ground at 16 m/s and 40° above the horizontal towards a wall 10 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 60° above horizontal from the top of a 35 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 10 kg block slides 20 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.35. Find the work Wf done against friction.
A 10 kg block starts from rest at the bottom of a 15° slope and is pulled up the slope by a constant force of 50 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 2 m up the slope.
A 1500 kg car moving at 19 m/s runs into a stationary 1200 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.4 kg ball hits a wall at 12 m/s and bounces straight back at 10 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water drains from a large open tank through a sharp-edged orifice with a diameter of 20 mm. The water level is kept 3.5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 150 mm to 50 mm while rising 1 m. At section 1 the pressure is 160 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 310 g of Ar occupy at 500 kPa and 95 °C? Treat it as an ideal gas.
A fixed amount of He goes from 250 kPa, 2 L, 20 °C to 200 kPa and 230 °C. Find the new volume V2.
A reversible refrigerator keeps a cold room at 5 °C by removing 7.5 kW of heat, and rejects heat to the surroundings at 28 °C. Find the rate Q̇H at which it rejects heat.
A refrigerator keeps its cold space at -10 °C in a kitchen at 28 °C, removing heat from the cold space at 5 kW. Find the minimum power input Ẇmin.
A string fixed at both ends is 0.65 m long; waves travel on it at 200 m/s. Find the frequency f5 of its harmonic number 5.
A sensor at rest sends ultrasound of 40 kHz towards a car that approaches at 30 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
Answer: d = 33.59 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Answer: t = 2.951 s
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 2.356 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Down the slope is positive.
Answer: T = 81.43 N
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: y = 5.126 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
The ball is still rising at the wall.
Answer: x = 38.8 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: Wf = 526 J
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
Answer: v = 2.039 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
Answer: ΔK = 120333 J
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
The lost energy goes into deformation, heat and sound.
Answer: Favg = 352 N
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: F = 40.47 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: F = 207.2 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: Q = 1.614 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: p2 = 140.2 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Work in Pa, then divide by 1000 to get kPa.
Answer: V = 47.5 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 4.291 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Q̇H = 8.12 kW
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Power input: Ẇ = 0.6202 kW.
Answer: Ẇmin = 0.722 kW
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
The least work is needed by a reversible machine, which has the highest COP.
Answer: f5 = 769.2 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Answer: fb = 7668 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
ExamProblems · examproblems.com/courses/physics-for-engineers