Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 72 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 8 m/s from the edge of a roof 5 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 10 kg box is pushed up a ramp at 40° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the pushing force F.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 130 N pulls the front 5 kg block; the 4 kg block is towed behind. The coefficient of kinetic friction is 0.2 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 65° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 30° above horizontal from the top of a 40 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A small ball on a string of length 0.6 m moves in a horizontal circle so that the string makes a constant angle of 40° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
A 1 kg cart is released from rest at a height of 13.5 m on a frictionless track and runs through a vertical loop of radius 3 m. Use g = 9.81 m/s². Find the normal force Ntop of the track on the cart at the top of the loop.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.2. Find its speed v after it has slid 19 m along the slope.
An 8 kg block rests on a rough horizontal table (μ = 0.15). A light string runs from it over a smooth pulley at the edge of the table to a hanging 5 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 0.5 m.
A 900 kg car moving at 19 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 1000 kg car travelling east at 22 m/s collides with a 1900 kg car travelling north at 14 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A shaft transmits 172.5 kW at 1200 rpm. Find the torque T in the shaft.
A hollow sphere starts from rest and rolls without slipping down a slope inclined at 20° to the horizontal. Use g = 9.81 m/s². Find its speed v after it has rolled 4 m along the slope.
A 15 kW motor running at 2900 rpm drives a pinion with 12 teeth, which meshes with a gear with 24 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
In a compound gear train, gear A (60 teeth) on the input shaft turns at 1440 rpm and drives gear B (80 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 720 rpm. How many teeth ND must gear D have?
A simple pendulum has a length of 1 m. Use g = 9.81 m/s². Find its period T for small swings.
A 1.1 kg mass on a spring with k = 120 N/m is pulled 5 cm from equilibrium and released. Find the total energy E of the oscillation.
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed one cycle later by a peak of 11.7 mm. Find the damping ratio ζ.
A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 25 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: t = 2.667 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.
Answer: hmax = 8.262 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: F = 85.6 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.
Constant speed: the forces balance.
Answer: T = 57.78 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: t = 6.282 s
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.
Starting and landing at the same height: t = 2vy/g.
Answer: x = 52.45 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: T = 1.36 s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: Ntop = 39.24 N
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Towards the centre is positive.
Answer: v = 9.485 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).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: v = 1.693 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/K = 66.67 %
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′ = 11.9 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: T = 1373 N·m
Rotation about a fixed axis: the sum of the moments about the axis equals I·α (ΣM = I·α), just as ΣF = m·a for translation. In energy form, a torque T turning through an angle θ (in radians) does work T·θ, which changes the kinetic energy ½·I·ω²; gravity adds or removes m·g·Δh of the centre of mass. Power is P = T·ω.
Answer: v = 4.013 m/s
Rolling and angular momentum: a body rolling without slipping has v = ω·r, so its kinetic energy is ½·m·v² + ½·I·ω² = ½·m·v²·(1 + I/(m·r²)). Down a slope that gives a = g·sin θ/(1 + I/(m·r²)). When no external moment acts about an axis, the angular momentum about that axis stays constant, I₁·ω₁ = I₂·ω₂, even if kinetic energy is lost.
Rolling without slipping: ω = v/r, and friction does no work at the contact point.
Answer: T1 = 49.39 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Answer: ND = 24
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Answer: T = 2.006 s
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: E = 0.15 J
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: ζ = 0.12
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Answer: keq = 8.108 kN/m
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
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