Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 45 km/h brakes with a constant deceleration of 3.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 12 m/s from the edge of a roof 35 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 12 kg block slides down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 1 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 14 m/s and 40° above the horizontal towards a wall 14 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 22 m/s and 50° above the horizontal from the top of a 20 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A 4 kg block slides 5 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 1 kg block is released from rest at height 2 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.1) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1900 kg car moving at 15 m/s runs into a stationary 1800 kg car and the two lock together. What percentage ΔK/K of the original kinetic energy is lost?
At a crossing a 1200 kg car travelling east at 14 m/s collides with a 1400 kg car travelling north at 10 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 4 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 seawater, with its top edge 1 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 40 mm. The water level is kept 6 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 1 m. At section 1 the pressure is 130 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 50 L contains 220 g of N₂ at 100 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CH₄ goes from 100 kPa, 10 L, 25 °C to 300 kPa and 50 °C. Find the new volume V2.
A heat engine receives heat from a source at 700 °C and rejects heat to the surroundings at 20 °C. Its actual thermal efficiency is 45%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat pump delivers 14.5 kW of heat to a heating system at 55 °C, taking heat from a source at 7 °C. Its actual COP is 50% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
A pipe closed at one end is 0.8 m long. The speed of sound is 343 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 500 nm passes through a single slit 0.08 mm wide. Find the width w of the central bright maximum on a screen 2 m away (between the first dark fringes on either side).
Answer: s = 22.32 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 = 4.161 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 = 5.554 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 = 10.03 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.
Answer: y = 3.389 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 already on its way down at the wall.
Answer: α = 61.47 °
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.
Energy or v² = v₀² + 2as in the vertical direction.
Answer: Wf = 45.09 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: x = 19.5 cm
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).
The ramp is smooth, so only the strip removes energy.
Answer: ΔK/K = 48.65 %
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: v′ = 8.411 m/s
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 = 88.29 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 = 113.1 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 = 8.59 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 = 95.41 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: P = 487.3 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 3.613 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 64.4 %
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.
Answer: Ẇ = 4.242 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.
Answer: f3 = 321.6 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: w = 25 mm
Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …
ExamProblems · examproblems.com/courses/physics-for-engineers