Edexcel IGCSE Physics: Double Science 复习笔记:1.2.4 Unbalanced Forces

Edexcel IGCSE Physics: Double Science 复习笔记:1.2.4 Unbalanced Forces

Unbalanced Forces

 

  • Forces can combine to produce
    • Balanced forces
    • Unbalanced forces

     

 

 

  • Balanced forces mean that the forces have combined in such a way that they cancel each other out and no resultant force acts on the body
    • For example, the weight of a book on a desk is balanced by the normal force of the desk
    • As a result, no resultant force is experienced by the book, the book and the table are equal and balanced

     

 

 

5.1.6-Balanced-Forces

 

A book resting on a table is an example of balanced forces

 

 

 

  • Unbalanced forces mean that the forces have combined in such a way that they do not cancel out completely and there is a resultant force on the object
    • For example, imagine two people playing a game of tug-of-war, working against each other on opposite sides of the rope
    • If person A pulls with 80 N to the left and person B pulls with 100 N to the right, these forces do not cancel each other out completely
    • Since person B pulled with more force than person A the forces will be unbalanced and the rope will experience a resultant force of 20 N to the right

     

 

5.1.6-Tug-O-War

 

A tug-of-war is an example of when forces can become unbalanced

 

 

 

Unbalanced Forces, Mass & Acceleration

  • When forces combine on an object in such a way that they do not cancel out, there is a resultant force on the object
  • This resultant force causes the object to accelerate (i.e. change its velocity)
    • The object might speed up
    • The object might slow down
    • The object might change direction

     

  • The relationship between resultant force, mass and acceleration is given by the equation:

 

F = m × a

  • Where:
    • F = resultant force, measured in Newtons (N)
    • m = mass, measured in kilograms (kg)
    • a = acceleration, measured in metres per second squared (m/s2)

     

 

Worked Example

A car salesman says that his best car has a mass of 900 kg and can accelerate from 0 to 27 m/s in 3 seconds.Calculate:

 

a) The acceleration of the car in the first 3 seconds.

b) The force required to produce this acceleration. 

 

 

Part (a)

 

Step 1: List the known quantities

 

    • Initial velocity = 0 m/s
    • Final velocity = 27 m/s
    • Time, t = 3 s

     

 

Step 2: Calculate the change in velocity

 

change in velocity = Δv = final velocity − initial velocity

 

Δv = 27 − 0 = 27 m/s

 

Step 3: State the equation for acceleration

 

acceleration-equation-symbols-1

 

Step 4: Calculate the acceleration

 

a = 27 ÷ 3 = 9 m/s2

 

 

Part (b)

 

Step 1: List the known quantities

 

    • Mass of the car, m = 900 kg
    • Acceleration, a = 9 m/s2

     

 

Step 2: Identify which law of motion to apply

 

    • The question involves quantities of forcemass and acceleration, so Newton's second law is required:

     

 

F = ma

 

Step 3: Calculate the force required to accelerate the car

 

F = 900 × 9 = 8100 N

 

Worked Example

A passenger of mass 70 kg travels in a car at a speed of 20 m/s.The vehicle is involved in a collision, which brings the car (and the passenger) to a halt in 0.1 seconds.Calculate:

 

a) The deceleration of the car (and the passenger).

b) The decelerating force on the passenger. 

 

 

Part (a)

 

Step 1: List the known quantities

 

    • Initial velocity, u = 20 m/s
    • Final velocity, v = 0 m/s
    • Time, t = 0.1 s

     

 

Step 2: Calculate the change in velocity of the car (and the passenger)

 

change in velocity = Δv = final velocity − initial velocity = v − u

 

Δv = 0 − 20

 

Δv = −20 m/s

 

Step 3: Calculate the deceleration of the car (and the passenger) using the equation:

 

acceleration-equation-symbols-1

 

Step 4: Calculate the deceleration

 

a = −20 ÷ 0.1

 

a = −200 m/s2

 

 

Part (b)

 

Step 1: List the known quantities

 

    • Mass of the passenger, m = 70 kg
    • Acceleration (deceleration, in this case), a = −200 m/s2

     

 

Step 2: State the relationship between resultant force, mass and acceleration

 

    • This question involves quantities of force, mass and acceleration, so the appropriate equation for this case is:

     

 

F = m × a

 

Step 3: Calculate the decelerating force

 

F = 70 × −200

 

F = −14 000 N

 

 

Exam Tip

Remember that resultant force is a vector quantityExaminers may ask you to comment on why its value is negative - this happens when the resultant force acts in the opposite direction to the object's motionIn the worked example above, the resultant force opposes the passenger's motion, slowing them down (decelerating them) to a halt, this is why it has a minus symbol.

转载自savemyexams

更多IGCSE课程
翰林国际教育资讯二维码