Newton’s second law: F = ma
Newton's second law and the equation F = ma, finding the net force from a force diagram, calculating force, mass and acceleration, mass and weight, investigating acceleration with a trolley and hanging masses, and terminal velocity.
An empty bus and a bus packed with 60 school students have the same engine. Pull away from a stop and the empty bus surges forwards, while the full one moves off sluggishly. Newton's second law tells you exactly how much slower.
Force, mass and acceleration
Newton's first law says what happens with no net force. The second law says what a net force does: it makes the object accelerate, in the direction of the net force.
- The bigger the net force, the bigger the acceleration (for the same mass). Double the force, double the acceleration.
- The bigger the mass, the smaller the acceleration (for the same force). Double the mass, halve the acceleration.
Newton's second law: net force = mass × acceleration
F = m a so a = F ÷ m and m = F ÷ a
With mass in kg and acceleration in m/s², force is in newtons. In fact, this is how the newton is defined: 1 N is the net force that gives a 1 kg mass an acceleration of 1 m/s². Remember that F is the net force: you must add up all the forces first.
Horizontally: 400 N forwards − 300 N backwards = 100 N forwards. Vertically: 900 N down and 900 N up cancel. Net force = 100 N forwards.
The weight is 900 N, so the mass of cyclist and bike = 900 ÷ 9.8 = 91.8 kg.
a = F ÷ m = 100 ÷ 91.8 = 1.1 m/s² forwards.
If thrust drops to 300 N, the net force is zero and the cyclist keeps a constant velocity. If they stop pedalling, the net force is 300 N backwards, giving a deceleration of 300 ÷ 91.8 = 3.3 m/s².
A 10 kg mass accelerates at 2 m/s². F = m a = 10 × 2 = 20 N.
A net force of 70 N gives an object an acceleration of 3.5 m/s². m = F ÷ a = 70 ÷ 3.5 = 20 kg.
A net force of 2 N acts on a 500 g ball. Convert: 500 g = 0.5 kg. a = 2 ÷ 0.5 = 4 m/s².
A 50 N push accelerates a 5 kg box along the floor at 2 m/s². The net force needed is 5 × 2 = 10 N, so friction = 50 − 10 = 40 N.
Mass and weight
Mass is the amount of matter in an object, measured in kilograms. It is the same everywhere. Weight is the force of gravity on an object, measured in newtons. It comes straight from F = ma, using the acceleration due to gravity:
weight = mass × g W = m g (g = 9.8 N/kg on Earth)
A 100 g apple has a weight of 0.1 × 9.8 = 0.98 N, about 1 N. On the Moon, where g is 1.6 N/kg, the same apple weighs only 0.16 N. Gravity never completely disappears, even in deep space, though it gets weaker with distance. Astronauts on the International Space Station float not because there is no gravity there (it is about 90% as strong as on the ground), but because they and the station are falling around the Earth together.
Investigating F = ma
A trolley on a smooth bench can be pulled by a string over a pulley to a hanging mass. The weight of the hanging mass is the pulling force. Because the hanging mass accelerates too, the total mass being accelerated is the trolley plus the hanging mass. A motion sensor, light gate or ticker timer measures the acceleration.
- Use a 1 kg trolley and a 200 g hanging mass. Check the sim's working: F = 0.2 × 9.8 = 1.96 N, total mass 1.2 kg, a = 1.96 ÷ 1.2 = 1.63 m/s².
- Keep the hanging mass at 200 g and record a result for every trolley mass. Plot a against total mass, then a against 1 ÷ mass. Which graph is a straight line?
- Clear the graph. Keep the trolley at 2.0 kg and record each hanging mass. Plot a against force. (The total mass changes slightly each time; how could you keep it constant? Hint: move masses from the trolley to the hanger.)
- Turn on Friction and try the 50 g hanging mass. Why doesn't the trolley move?
Falling with air resistance
In Year 7 you saw that a skydiver reaches a steady terminal velocity. Newton's second law explains why. Just after jumping, the only force is weight, so a = g. As the skydiver speeds up, air resistance grows, so the net force (weight − air resistance) shrinks and so does the acceleration. At about 200 km/h in a belly-down position, air resistance equals weight: the net force is zero and the velocity stays constant. Opening the parachute suddenly makes air resistance much bigger than weight, so the net force is upwards and the skydiver decelerates to a new, much slower terminal velocity of about 20 km/h.
Test yourself
A net force of 30 N acts on a 6 kg trolley. Its acceleration is
a = F ÷ m = 30 ÷ 6 = 5 m/s².
The same net force acts on an empty bus and on the same bus full of passengers. Compared with the empty bus, the full bus
a = F ÷ m; more mass means less acceleration.
A 1200 kg car accelerates at 2.5 m/s². The net force on it is
F = m a = 1200 × 2.5 = 3000 N.
A 60 kg student's weight on Earth is about
W = m g = 60 × 9.8 = 588 N.
A driver lifts their foot until the car travels at a constant 80 km/h. The net force on the car is
Constant velocity means zero acceleration, so F = m × 0 = 0.
Label the forces on the cyclist.
Order these from smallest to largest acceleration (each uses a net force of 12 N).
Newton's second law states that net force equals mass × . Force is measured in . Doubling the mass while keeping the force the same will the acceleration. Weight is mass × .
A cyclist and bike have a total mass of 95 kg. The cyclist applies a net force of 25 N for 50 s, starting from rest. (a) Calculate the acceleration and the top speed reached. (b) The cyclist then stops pedalling and the drag and friction total 20 N. Calculate the deceleration and explain, using Newton's laws, what happens to the bike's motion.
(a) a = F ÷ m = 25 ÷ 95 = 0.26 m/s². Starting from rest, v = u + a t = 0 + 0.263 × 50 = 13 m/s, which is about 47 km/h. (b) When the cyclist stops pedalling there is no forward force, so the net force is the 20 N of drag and friction acting backwards. By Newton's second law the deceleration is a = F ÷ m = 20 ÷ 95 = 0.21 m/s². The bike does not stop immediately: its inertia (first law) keeps it moving, and the backward net force reduces its speed by about 0.21 m/s every second. As it slows, air resistance decreases, so the deceleration gets smaller, until friction finally brings it to rest.
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