How Forces Affect Motion · Lesson 5 of 8
Newton’s Second Law of Motion
“The second law turns cause and response into a calculation linking net force, mass, and acceleration.”
• Relate acceleration directly to net force and inversely to mass. • Apply the mathematical forms of Newton’s second law. • Define one newton. • Calculate gravitational force near the Earth. • Combine force calculations with motion equations and graphs. • Explain how increasing stopping time reduces force.
A gentle push gives an empty cart a modest change in velocity. A stronger push changes its velocity more quickly. Load the cart with books and the same push becomes less effective. These observations reveal two independent relationships: acceleration grows with net force, but decreases as mass grows.
Force And Acceleration
For the same object, increasing the net force increases acceleration in the same ratio under ideal conditions. Doubling the net force doubles acceleration; halving the net force halves acceleration. Direction matters as well: acceleration points in the direction of the net force. If the net force points opposite the velocity, the object slows.
Mass And Acceleration
For the same net force, a smaller mass has a larger acceleration and a larger mass has a smaller acceleration. Doubling mass halves acceleration when net force remains fixed. Mass therefore measures resistance to acceleration and is closely connected with inertia.
When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of acceleration is proportional to net force and inversely proportional to mass.
Definition Of One Newton
One newton is the force that produces an acceleration of one metre per second squared in an object of mass one kilogram.
Problem
A net force of 18 N acts on a 6 kg trolley. Find its acceleration.
- 1.Write the known quantities: Fₙₑₜ = 18 N and m = 6 kg.
- 2.Choose a = Fₙₑₜ/m.
- 3.Substitute: a = 18 N ÷ 6 kg.
- 4.Calculate: a = 3 m s⁻².
- 5.State direction: the acceleration is 3 m s⁻² in the direction of the net force.
Problem
A 1200 kg car accelerates east at 2.5 m s⁻². Find the net force.
- 1.Write m = 1200 kg and a = 2.5 m s⁻² east.
- 2.Use Fₙₑₜ = ma.
- 3.Substitute: Fₙₑₜ = 1200 × 2.5.
- 4.Calculate: Fₙₑₜ = 3000 N.
- 5.Attach direction: 3000 N towards the east.
Gravitational Force Near The Earth
An object released near the Earth accelerates downward because the Earth exerts gravitational force on it. The acceleration is denoted by g and is about 9.8 m s⁻² near the surface. For quick estimation it may be taken as 10 m s⁻². Substituting g for acceleration in the second law gives the object’s weight.
Problem
A barbell has two 10 kg plates and a 10 kg bar. Find the upward force required to hold it steady using g = 9.8 m s⁻².
- 1.Find total mass: 10 kg + 10 kg + 10 kg = 30 kg.
- 2.Calculate weight: Fᵍ = mg = 30 × 9.8 = 294 N downward.
- 3.Holding steady means acceleration is zero and forces balance.
- 4.The upward force must equal the downward weight.
- 5.The required force is 294 N upward.
Working With Friction
The applied force is not automatically the force used in F = ma. First calculate the net force. If a person pushes right with 55 N and friction acts left with 50 N, then the net force is only 5 N right. This smaller result determines acceleration.
Problem
A 25 kg block starts from rest. It is pushed right with 55 N while friction is 50 N left. Find its displacement in 2 s.
- 1.Calculate net force: 55 N − 50 N = 5 N right.
- 2.Use the second law: a = Fₙₑₜ/m = 5/25 = 0.2 m s⁻² right.
- 3.Write the initial velocity u = 0, time t = 2 s, and acceleration a = 0.2 m s⁻².
- 4.Use s = ut + ½at².
- 5.Substitute: s = 0 × 2 + ½ × 0.2 × 2².
- 6.Calculate: s = 0.4 m.
- 7.The displacement is 0.4 m towards the right.
Activity: Changing Force
Use a low-friction cart pulled by a thread passing over a pulley to a hanging cup. Keep the cart mass fixed and increase the hanging mass to increase the pulling force. Release from the same position and measure the time to cover the same distance. With initial velocity zero, s = ½aT². A shorter travel time for the same distance indicates a larger acceleration.
Activity: Changing Mass
Keep the hanging cup and its load unchanged so that the pulling force remains approximately fixed. Add mass to the cart and repeat the run over the same distance. The heavier cart takes longer and has smaller acceleration. Friction and measurement uncertainty may keep the ratios from being exact, but the pattern should remain.
Momentum Form Of The Law
Momentum is the product of mass and velocity, and its direction is the direction of velocity.
A more general form of the second law says that net force is related to the rate of change of momentum and acts in the direction of that change. This form is useful even when mass is not constant. For constant mass, it leads to F = ma.
Increasing Stopping Time
A fast ball must change from high velocity to zero when caught. If the change happens in a very short time, the acceleration magnitude is large and so is the force. A fielder moves the hands backwards with the ball, increasing the stopping time and reducing acceleration. Airbags, landing mats, bubble wrap, and hay use the same principle: they extend the time or distance over which velocity changes.
A coconut striking a hard floor stops in a very short time. The large acceleration requires a large force, which can crack the shell. A soft cushion deforms and allows the stop to occur more gradually, reducing the force.
Reading A Velocity-Time Graph
The slope of a velocity-time graph gives acceleration. An upward straight segment has constant positive acceleration; a horizontal segment has zero acceleration; a downward straight segment has constant negative acceleration. Once acceleration is obtained for each interval, multiply by mass to calculate net force.
Problem
A 1500 kg car increases velocity from 0 to 10 m s⁻¹ in 5 s, moves at 10 m s⁻¹ for 5 s, then stops uniformly in 5 s. Find the force in each interval.
- 1.First interval: a = (10 − 0)/5 = 2 m s⁻². Therefore F = 1500 × 2 = 3000 N in the direction of motion.
- 2.Middle interval: velocity is constant, so a = 0 and Fₙₑₜ = 0 N.
- 3.Final interval: a = (0 − 10)/5 = −2 m s⁻². Therefore F = 1500 × (−2) = −3000 N.
- 4.Interpret the negative sign: the final force has magnitude 3000 N opposite to the original direction of motion.
• Draw or describe all relevant forces. • Calculate the directed net force first. • Convert mass to kilograms and time to seconds. • Select F = ma or a = F/m. • Combine with a motion equation only after acceleration is known. • State units and direction. • Interpret negative signs as direction.
Quiz
Which description best matches Newton’s Second Law Of Motion?
Which description best matches One Newton?
Which term matches this description: When a net force acts on an object, the object accelerates in the direction of the net force.
Which term matches this description: One newton is the force that produces an acceleration of one metre per second squared in an object of mass one kilogram.
Which statement is a key takeaway from this lesson?
Practice Problems
- A 100 g toy car moves with constant velocity. Determine its net force.
- Explain why the same push accelerates an empty trolley more than a loaded trolley.
- Calculate the weight of a 12 kg object using g = 9.8 m s⁻².
- Explain why glass objects are protected by bubble wrap during transport.
Key Takeaways
• Acceleration is directly proportional to net force and inversely proportional to mass. • Acceleration points in the direction of net force. • The equation uses net force: Fₙₑₜ = ma. • One newton equals one kilogram metre per second squared. • Weight near the Earth is mg. • Increasing stopping time reduces acceleration and force. • Velocity-time slope gives acceleration.
Forces do not occur in isolation. The next lesson examines the equal and opposite forces created whenever two objects interact.