Dynamics and Conservation Laws
Study of the causes of motion: forces, energy, work. Newton's laws and conservation of mechanical energy.
Complete Theory
6A body with zero net force remains at rest or continues in uniform linear motion (ULM). This is Newton's First Law of Inertia: a body does not change its state of motion unless compelled to by a net external force.
Inertial mass is the property of matter that resists changes in velocity — a heavier object requires a greater force to achieve the same acceleration.
An inertial frame is a reference frame in which the First Law is valid: a body subject to no forces moves in a straight line at constant speed. Any frame in uniform translation relative to an inertial frame is itself inertial. The Earth is a good approximation of an inertial frame for most laboratory purposes (the corrections due to its rotation are negligible except for large-scale atmospheric phenomena).
Inertial mass is the property of matter that resists changes in velocity — a heavier object requires a greater force to achieve the same acceleration.
An inertial frame is a reference frame in which the First Law is valid: a body subject to no forces moves in a straight line at constant speed. Any frame in uniform translation relative to an inertial frame is itself inertial. The Earth is a good approximation of an inertial frame for most laboratory purposes (the corrections due to its rotation are negligible except for large-scale atmospheric phenomena).
The net (resultant) force acting on a body equals the product of its mass and its acceleration: . This is the quantitative heart of Newtonian mechanics.
Inertial mass measures the resistance of a body to changes in its state of motion. A larger mass requires a proportionally larger force to produce the same acceleration.
In practice the law is applied component by component: , . A free-body diagram is drawn first to identify all forces; the net force in each direction is then set equal to in that direction. If the net force is zero the body is in dynamic equilibrium (First Law).
Inertial mass measures the resistance of a body to changes in its state of motion. A larger mass requires a proportionally larger force to produce the same acceleration.
In practice the law is applied component by component: , . A free-body diagram is drawn first to identify all forces; the net force in each direction is then set equal to in that direction. If the net force is zero the body is in dynamic equilibrium (First Law).
Forces always come in pairs: if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. The two forces are equal in magnitude and opposite in direction.
A crucial point: the action and reaction forces act on different bodies — they can never cancel each other in a single free-body diagram. For example, when you push against a wall the wall pushes back on you with the same force; these two forces act on different objects (wall vs. you) and do not cancel.
The Third Law is the microscopic origin of the conservation of momentum: for an isolated system the internal action–reaction pairs sum to zero, so the total momentum is constant.
A crucial point: the action and reaction forces act on different bodies — they can never cancel each other in a single free-body diagram. For example, when you push against a wall the wall pushes back on you with the same force; these two forces act on different objects (wall vs. you) and do not cancel.
The Third Law is the microscopic origin of the conservation of momentum: for an isolated system the internal action–reaction pairs sum to zero, so the total momentum is constant.
Weight always acts vertically downward toward the Earth's centre. On an inclined plane at angle it is convenient to decompose weight along the plane: (tends to slide the body down) and perpendicular to the plane: (balanced by the normal force).
The normal force is the contact force perpendicular to the surface; on a horizontal surface , on an incline .
Friction opposes relative motion or its tendency:
The normal force is the contact force perpendicular to the surface; on a horizontal surface , on an incline .
Friction opposes relative motion or its tendency:
- Static friction : prevents motion; it takes any value from zero up to its maximum .
- Kinetic (dynamic) friction : constant opposing force once sliding has begun; always .
The work done by a force along a displacement is , where is the angle between force and displacement. Only the component of force along the displacement does work; a force perpendicular to the motion (e.g. the normal force, centripetal force) does zero work.
The Work–Energy Theorem states that the net work done on a body equals its change in kinetic energy: . This is a scalar relation derived directly from Newton's Second Law integrated over the path.
A force is conservative if the work it does is independent of the path taken between two points (equivalently, the work around any closed path is zero). Gravity and the spring force are conservative; friction is not. For conservative forces a potential energy function exists such that .
Power is the rate at which work is done [W = J/s].
The Work–Energy Theorem states that the net work done on a body equals its change in kinetic energy: . This is a scalar relation derived directly from Newton's Second Law integrated over the path.
A force is conservative if the work it does is independent of the path taken between two points (equivalently, the work around any closed path is zero). Gravity and the spring force are conservative; friction is not. For conservative forces a potential energy function exists such that .
Power is the rate at which work is done [W = J/s].
When only conservative forces act, the total mechanical energy is constant: . This is the law of conservation of mechanical energy. Energy is continuously exchanged between kinetic and potential forms, but their sum never changes.
Two important potential energy functions:
When non-conservative forces (such as friction) are present, the work they do equals the change in mechanical energy: . Friction converts mechanical energy irreversibly into heat, so decreases.
Two important potential energy functions:
- Gravitational PE (near Earth's surface): , measured from a chosen reference height .
- Elastic PE (spring): , where is the compression or extension from the natural length and is the spring constant [N/m].
When non-conservative forces (such as friction) are present, the work they do equals the change in mechanical energy: . Friction converts mechanical energy irreversibly into heat, so decreases.
Worked Examples
3Example 1Block on inclined plane with friction
Given
block mass
incline angle
coefficient of kinetic friction
length of the incline
Find
Acceleration of the block along the incline
Speed after travelling 4 m (starting from rest)
Step-by-step solution
1Setup: a 5 kg block slides down an incline of 30° with friction. The block starts from rest (). The forces acting are: weight (vertical), normal reaction (perpendicular to the incline), and friction (parallel to the incline, opposite to motion).
2Weight decomposition: the weight is resolved into two components: (parallel to the incline, downward, the component that "pulls" the block) and (perpendicular to the incline).
3Normal reaction and friction: the perpendicular component of the weight is balanced by the normal reaction: . The kinetic friction force is proportional to : , directed up the incline (opposes the descent).
4Application of Newton's second law: along the incline, the net force is (downward). The acceleration is . Without friction it would be .
5Speed after 4 m: the motion along the incline is UAM with constant acceleration . Using with : . Alternatively, using energy: , giving .
6Energy check: potential energy lost: . Work done by friction: . Kinetic energy gained: . Note: .
✓ Final result: ,
Example 2Atwood machine
Given
larger mass (left)
smaller mass (right)
Ideal pulley (massless, frictionless, inextensible string)
Find
Acceleration of the system
Tension in the string
Step-by-step solution
1Setup: two masses and are connected by an inextensible string that passes over an ideal pulley. The larger mass descends, the smaller rises, both with the same acceleration magnitude (the string is inextensible). Pulley mass and friction are neglected.
2Force diagram: on act downward and upward. On act downward and upward. Choose the positive direction for as "downward for " (and therefore "upward for ").
3Equations of motion: for : . For : . Adding the two equations eliminates : , hence .
4Tension: substitute into one of the equations, e.g. . Note that the tension lies between (weight of the smaller mass) and (weight of the larger mass).
5Check limiting cases: if , then and (equilibrium). If , then ( falls almost freely) and (the string is almost slack).
6Energy check: the potential energy lost by in time is . That gained by is . The difference transforms into kinetic energy of both masses: .
✓ Final result: ,
Example 3Compressed spring — launch speed
Given
spring constant
initial compression (8 cm)
block mass
Find
Speed of the block when the spring returns to its natural length
Step-by-step solution
1Setup: a spring compressed by 8 cm () launches a 200 g block on a smooth horizontal surface (no friction). The spring+block system is isolated and conservative: the elastic potential energy is completely converted into kinetic energy of the block.
2Elastic potential energy stored: . This is the energy stored in the spring when compressed by 8 cm.
3Conservation of energy: when released, the elastic potential energy transforms into kinetic energy of the block (the spring is ideal and massless). At the equilibrium position (): , hence .
4Dynamics check: alternatively, integrate . The motion is harmonic: with . The maximum speed occurs at and is .
5Comment: if the surface had friction, part of the energy would be dissipated as heat and the final speed would be lower. Conservation of mechanical energy holds only in the presence of conservative forces.
✓ Final result:
Exercises with Solutions
3Exercise 1Inclined plane — equilibrium?Hard
Problem to solve
A block of mass rests on an incline at . The coefficient of static friction is , kinetic friction . Determine: (a) whether the block remains at rest or slides, (b) if it slides, the acceleration, (c) the distance travelled after 2 seconds starting from rest.
Given data
m = 8\,kg (block mass)\theta = 35° (incline angle)\mu_s = 0.45 (static friction)\mu_k = 0.35 (kinetic friction)
Step-by-step solution
1Equilibrium check: the force tending to make the block slide is . The maximum static friction force is . Since , the block slides.
2Acceleration: once the block moves, friction becomes kinetic: (smaller than maximum static friction). The net force is . The acceleration is .
3Distance after 2 seconds: the motion is UAM with : . After 2 seconds, the block has travelled 5.64 m along the incline.
4Critical angle: the angle beyond which the block begins to slide is . For , the block stays at rest; for , it slides.
✓ Final answer: Not in equilibrium. ,
Exercise 2Bullet embedding in pendulumVery Hard
Problem to solve
A bullet of mass (20 g) travels at and embeds itself in a wooden block of mass suspended by a string of length (ballistic pendulum). Determine: (a) the speed of the block+bullet immediately after the collision, (b) the maximum height reached by the pendulum, (c) the tension in the string right after the collision (at the lowest point).
Given data
m_p = 0.02\,kg (bullet mass)v_0 = 300\,m/s (bullet speed)M = 2\,kg (block mass)L = 0.8\,m (string length)
Step-by-step solution
1Setup: the problem involves two distinct phases: (1) a perfectly inelastic collision between bullet and block (momentum conserved, energy not), (2) the pendulum swing (mechanical energy conserved after the collision).
2(a) Speed after collision: in a perfectly inelastic collision, bullet and block stick together. By conservation of momentum: , hence .
3(b) Maximum height: after the collision, the block+bullet system rises like a pendulum. By conservation of mechanical energy (friction neglected): , hence (about 45 cm). The corresponding swing angle is .
4(c) Tension after collision: right after the collision, when the pendulum is still at the lowest point, the string tension must support the weight and provide the centripetal force for circular motion: . Hence .
5Comment: the ballistic pendulum is a historical device for measuring projectile speed: by measuring you obtain , and from you recover .
✓ Final answer: , ,
Exercise 3Road curveHard
Problem to solve
A car of mass travels around a circular curve of radius at speed (72 km/h). The coefficient of static friction between tyres and asphalt is . (a) Is friction sufficient to keep the car on the curve? If not, what is the maximum safe speed? (b) If the road is banked at an angle , what is the ideal speed (no friction needed)?
Given data
m = 1200\,kg (car mass)r = 80\,m (curve radius)v = 20\,m/s (speed)\mu_s = 0.5 (static friction)
Step-by-step solution
1Setup: on a curve, the car has centripetal acceleration directed toward the centre. The force providing this acceleration (centripetal force) is static friction between tyres and asphalt (for a flat, unbanked curve).
2(a) Check: the required centripetal force is . The maximum available static friction is . Since , friction is not sufficient: the car would skid outward.
3Maximum speed: the maximum safe speed is found by equating , i.e. (about 71 km/h). The actual speed of 20 m/s (72 km/h) is just above the limit.
4(b) Banked curve: if the road is tilted inward by , a component of the normal reaction contributes to the centripetal force. At the ideal speed, friction is not needed: and . Dividing gives , hence (about 52 km/h).
5Comparison: the banking allows the curve to be taken at higher speeds safely, because part of the centripetal force is provided by the horizontal component of the normal reaction, reducing the demand on friction.
✓ Final answer: (a) Friction insufficient; . (b) for .
Keep studying
Guided exercises on this topic
Recommended Books
Introductory
Physics for Scientists and Engineers
Buy on Amazon →
Advanced
Classical Mechanics
Buy on Amazon →
As an Amazon Associate I earn from qualifying purchases.
🔥
Integrative Problems
Problems combining all chapters — exam levelProblem 1Tower, Ballistic Pendulum, and Keplerian OrbitEXTREME
A cannon is placed on top of a tower tall and fires a projectile of horizontally at .
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
The projectile strikes and embeds in a wooden block hanging from a rope of length (ballistic pendulum), at ground level.
The Earth-Moon system is then used as a reference for Kepler's third law.
📌 Problem data
(a)Uniformly Accelerated Motion(b)Inelastic Collision(c)Potential Energy + Pendulum(d)Moment of Inertia — Rigid Body(e)Gravitation — Kepler's Third Law
Problem 2Spring, Rolling Disk, Inclined Plane Collision, and ConservationEXTREME
A spring (, compressed ) launches a solid disk (, ) up an inclined plane (, , ) that rolls without slipping.
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
At the top the disk is launched horizontally and strikes a pendulum (, ) — perfectly inelastic collision. What is asked (solved below, a→e): (a) the disk's speed at the top of the plane; (b) the range and impact speed of the horizontal launch; (c) the speed after the inelastic collision with the pendulum and the energy lost; (d) the pendulum's maximum angle, the maximum tension, and whether it completes the loop; (e) the full energy balance (from spring to maximum angle).
📌 Problem data
(a)Energy + Rigid Body (rolling)(b)Kinematics — Projectile(c)Inelastic Collision + CM(d)Pendulum Dynamics + Forces(e)Conservation Laws — Complete Energy Balance