CIE M1T5 -Energy, work and power

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MECHANICS FOR PP4

M1 TOPIC 4 : Newton's Laws of Motion

Learning Objectives

Candidates should be able to:

Apply Newton's laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces
Use the relationship between mass and weight
Solve simple problems involving motion of a particle moving vertically or on an inclined plane with constant acceleration
Solve simple problems involving the motion of connected particles

Core Concepts & Definitions

Newton's First Law (Law of Inertia)

A body will remain at rest or continue to move with constant velocity unless acted upon by a resultant external force.

Newton's Second Law (F = ma)

The rate of change of momentum of a body is proportional to the resultant force acting on it and occurs in the direction of that force.

∑F = ma

Where ∑F is the resultant force (in Newtons, N), m is the mass (in kilograms, kg), and a is the acceleration (in m s⁻²).

Newton's Third Law (Action-Reaction)

If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.

Mass vs. Weight

Mass (m)

A measure of the amount of matter in an object (scalar, in kg). It is constant.

Weight (W)

The force acting on an object due to gravity (vector, in N). It depends on the gravitational field strength.

W = mg

Where g is the acceleration due to gravity. For this syllabus, use the approximate value g = 10 m s⁻².

Common Forces in Problems

Weight (W or mg)

Always acts vertically downwards.

Normal Reaction (R or N)

The force exerted by a surface on an object. It acts perpendicular to the surface.

Friction (F or F_r)

Acts parallel to the surface, opposing (or preventing) motion.

Maximum Static Friction: F ≤ μR
Kinetic (Dynamic) Friction: F = μR (when the object is moving)

Where μ is the coefficient of friction.

Tension (T)

The pulling force in a string, rope, or cable. In a light, inextensible string:

Tension is constant throughout its length
It does not sag or stretch

Thrust

The pushing force in a rigid connecting rod or tow-bar.

Problem-Solving Strategy

Draw a Diagram: Sketch the situation.
Model the Object: Represent it as a particle (so we can ignore its size and rotation).
Draw a Free-Body Diagram: Isolate the particle and draw all the forces acting on it as arrows.
Resolve Forces: Choose a sensible direction to resolve forces (typically parallel and perpendicular to the plane of motion).
Apply Newton's Second Law: In each direction, apply ∑F = ma.
Solve the Equations: Use the equations to find the unknown quantity.

Application to Specific Scenarios

A. Motion on an Inclined Plane

Resolve forces parallel and perpendicular to the plane
The weight (mg) must be resolved into components:
- Component parallel to plane: mg sin θ
- Component perpendicular to plane: mg cos θ
Key Insight: The acceleration up the plane and down the plane will be different if friction is present
- Moving up the plane: Friction acts down the plane
- Moving down the plane: Friction acts up the plane

B. Connected Particles

These problems involve applying F=ma to each particle separately and then solving the resulting system of equations.

Over a Smooth Pulley

Two particles connected by a light, inextensible string over a fixed pulley
The system accelerates in the direction of the heavier particle
The tension (T) is the same on both sides of the pulley
The magnitude of acceleration (a) is the same for both particles
Method:
1. Treat each particle separately
2. Apply F=ma to each
3. Solve the simultaneous equations for a and T

Towing (Trailer Problems)

A car towing a trailer using a light rope or a rigid tow-bar
If the link is a rope, it can only experience tension
If the link is a rigid tow-bar, it can experience both tension and thrust (compression)
Method:
1. Consider the whole system to find the acceleration
2. Then, isolate one part (e.g., just the trailer) to find the tension or thrust in the connecting link

Important Notes from the Syllabus

Other Forces

If a problem involves forces like air resistance, it will be explicitly stated. In standard problems, it is not considered.

Numerical Values

You are expected to use g = 10 m s⁻² for calculations unless stated otherwise.

Key Example

A classic problem is a particle on a rough inclined plane where you are asked to find the acceleration when it is pushed up the plane and compare it to the acceleration when it slides back down. The equations for ∑F will be different in each case due to the direction of friction.

Additional Required Concepts

Work-Energy Principle

Work done = Force × Distance × cosθ

Kinetic Energy = ½mv²

Potential Energy = mgh

Work-Energy Theorem: Work done by all forces = Change in kinetic energy

Vertical Projectile Motion

v = u + at

s = ut + ½at²

v² = u² + 2as

For vertical motion under gravity: a = -g = -10 m/s²

Limiting Equilibrium Analysis

For objects on the point of moving:

Friction = μR (maximum value)
Resultant force = 0 in all directions
Consider both up and down plane scenarios

Forces on a body on an inclined plane

Consider a body of mass m placed on a smooth inclined plane that makes an angle θ with the horizontal.

Weight (W) acts vertically downward. W = m g
Normal reaction (R) acts perpendicular to the plane.
Component of weight down the plane W_‖ = m g sinθ
Component of weight perpendicular to the plane W_⊥ = m g cosθ

When the body rests on the plane

The normal reaction balances the perpendicular component of weight:

R = m g cosθ

When the body slides down the plane

The net force down the plane is given by:

F = m g sinθ − f

where f is the frictional force (f = μR).

When the body is pulled up the plane

If a force P acts up the plane at angle α to the plane, the forces are:

Component of pull up the plane: P cosα
Component of pull perpendicular to plane: P sinα

Normal reaction: R = m g cosθ − P sinα

Net force up the plane: F = P cosα − (m g sinθ + f)

If motion is uniform (constant speed)

Then net force along the plane is zero:

P cosα = m g sinθ + f

Summary of key equations

W = m g
R = m g cosθ
f = μR
F_down = m g sinθ − f
F_up = P cosα − (m g sinθ + f)