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 8 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 19 m/s from the edge of a roof 20 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.
An 18 kg block slides down a slope at 20° 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.
A 2 kg block on a ramp at 40° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 1.5 kg mass. The hanging mass moves down. The ramp is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 38 m/s and 25° above the horizontal towards a wall 34 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 31 m/s and 55° above the horizontal from the top of a 25 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 1.9 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.3. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A 0.2 kg ball on a light string 0.6 m long is swung in a vertical circle. At the lowest point its speed is 12 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 0.2 kg ball hangs on a light string of length 0.5 m. It is pulled aside until the string makes 70° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 0.5 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.25) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1900 kg car moving at 21 m/s runs into a stationary 900 kg car and the two lock together. What percentage ΔK/K of the original kinetic energy is lost?
A 0.1 kg ball hits a wall at 25 m/s and bounces straight back at 13 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A shaft transmits 166.5 kW at 800 rpm. Find the torque T in the shaft.
A person stands on a frictionless turntable and holds a 2 kg dumbbell in each hand, 0.75 m from the axis. The person and turntable alone have a moment of inertia of 2 kg·m² about the axis. They turn at 45 rpm, then pull the dumbbells in to 0.35 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A belt runs on a 150 mm driving pulley at 1450 rpm. The tension is 400 N on the tight side and 80 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (18 teeth) on the input shaft turning at 2900 rpm drives gear B (30 teeth). Gear C (50 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
A simple pendulum has a length of 1.1 m. Use g = 9.81 m/s². Find its period T for small swings.
A 2.9 kg mass is attached to the lower end of an unstretched vertical spring with k = 340 N/m and released from rest. Use g = 9.81 m/s². Find the greatest extension xmax of the spring in cm.
An 8 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 15 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: s = 9.766 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 = 4.735 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 = 0.5897 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: a = 0.601 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.
Answer: y = 11.07 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: H = 57.87 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.
Answer: ω = 4.149 rad/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.
Answer: Ttop = 38.19 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: T = 4.544 N
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). On a circular path the net force towards the centre must equal mv²/r.
Check: T = mg(3 − 2cos θ) = 4.544 N.
Answer: x = 7.177 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 = 32.14 %
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 = 152 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: T = 1987 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: n2 = 76.81 rpm
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.
The kinetic energy increases: the extra energy comes from the work the person does pulling the dumbbells in.
Answer: P = 3.644 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: nout = 1450 rpm
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.
total ratio = 2 : 1
Answer: T = 2.104 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: xmax = 16.73 cm
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: fd = 15.41 Hz
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 = 65 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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