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Forces and Newton's Laws of Motion 4.2 Newton's Second Law of Motion  F = ma and calculations

Doc Brown's Physics exam study revision notes

Examples of worked out calculation practice exam questions and problem solving based on Newton's 2nd Law of Motion

INDEX for physics notes on Newton's Laws of Motion: concepts, formulae, calculations and problem solving

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4.2 Newton's Second Law of Motion  F = ma and calculations

What is Newton's second law of motion?

How do we do Newton's 2nd law calculations?

How do you use the equation F = ma in problem solving?

Newton's Second Law of Motion can be expressed as the acceleration of a body is directly proportional to the force applied to that body. Remember that acceleration is the change in velocity over a specified time period.

(i) The greater the resultant force acting on a body, the greater the acceleration of that body.

In fact, you can see from the equation for Newton's 2nd law that the acceleration experienced by an object is directly proportional to the force acting on it.     F A

(ii) The acceleration of a body is also inversely proportional to its mass:   a 1/m

Those two statements can be expressed in the general formula:

resultant force (newtons) = mass (kilograms) x acceleration (metres per seconds squared),

this equation is a mathematical expression of Newton's 2nd law

F (N) = m (kg) x a (m/s2)

rearranging (i)  F = ma   gives  (ii)  a = F ÷ m     and   (iii)  m = F ÷ a

The greater the force acting on an object, the greater the object is accelerated.

Acceleration is inversely proportional to the mass.

If the same force is applied to objects of different mass, the smallest mass will be accelerated the most.

The consequence of statement (i) is that the faster you want to speed up (accelerate) an object the greater the force required (putting your foot down further on a car accelerator to increase the power). Conversely, the faster you want an object to slow down (decelerate) the greater the resistive force you must apply (pressing the brake pedal of a car more forcefully).

One consequence of statement (ii) is that a specific applied resultant force accelerates a body of smaller mass more than a body of larger mass. Or to express it another way, you need to apply a greater force to accelerate a larger body to the same extent as a smaller body.

Examples of calculations based on Newton's 2nd Law 

(some questions also need the formula for acceleration a = ∆v / ∆t)

Q2.1 What resultant force must be applied by a cyclist (mass 70.0 kg) to the pedals of a bike (8.50 kg) to give an acceleration of 0.15 m/s2?

Worked out ANSWERS

 

Q2.2 The engine of 1000 kg car generates a driving force of 4200 N.

If the car is travelling at 60 mph and the total resistive force (friction in engine, road contact and air resistance) is 3800 N what is the car's acceleration at this point?

Worked out ANSWERS

 

Q2.3 If a body experiences a resultant force of 500 N and accelerates away at 2.0 m/s2, what is the mass of the body?

Worked out ANSWERS

 

Q2.4 A bus of mass 2800 kg uniformly accelerates from 0 to 15 m/s in 20 seconds.

(a) Calculate the acceleration of the bus.

(b) Calculate the resultant force acting on the bus.

Worked out ANSWERS

 

Q2.5 The engine of a van of mass 2000 kg generates a driving force of 6000 N.

If it experiences a drag force due to air resistance of 500 N, calculate its acceleration.

Worked out ANSWERS

 

Q2.6 What force is required to vertically accelerate a 8000 kg rocket at 3 m/s2? (g = 9.8 m/s2)

Worked out ANSWERS

 

Q2.7 If the force exerted on an object is quadrupled and the mass doubled, what happens to the acceleration?

Worked out ANSWERS

 

Q2.8 High speed photography can be used to analyses the motion of sports equipment.

In cricket, a fast bowler sends the cricket ball down at 36 m/s (~80 mph, ~130 km/h) towards the batsman.

The cricket ball has a mass of 160 g and is a bit worn and slightly soft.

The batsman holds the bat still and blocks the ball which rebounds directly back towards the bowler at 30 m/s.

(a) If the contact time was found to be 0.002 seconds, what force did both the cricket ball experience and in what direction?

(i) You first need to calculate the acceleration

(ii) You can now calculate the force involved from Newton's 2nd Law equation.

(b) A new ball is harder than a used ball and the contact time may only be 0.001 seconds.

What effect does this have on the force experienced by a new cricket ball compared to the worn ball in (a)?

(c) Why might the impact time be longer for the worn ball compared to a new ball?

Worked out ANSWERS

 

INDEX for physics notes on Newton's Laws of Motion

Keywords, phrases and learning objectives for Newton's 2nd law of motion

Be able to know and use Newton's Second Law of Motion.

Know that acceleration is directly proportional to force applied to an object of a given mass and be able to solve problems using the formula F = ma with the appropriate units.

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Worked out ANSWERS to calculations using Newton's 2nd Law

Q2.1 What resultant force must be applied by a cyclist (mass 70.0 kg) to the pedals of a bike (8.50 kg) to give an acceleration of 0.15 m/s2?

F (N) = m (kg) x a (m/s2)

F = (70 + 8.5) x 0.15

force required = 11.8 N

 

Q2.2 The engine of 1000 kg car generates a driving force of 4200 N.

If the car is travelling at 60 mph and the total resistive force (friction in engine, road contact and air resistance) is 3800 N what is the car's acceleration at this point?

F (N) = m (kg) x a (m/s2)

The resultant forward force = 4200 - 3800 = 400 N

F = ma, so a = F / m = 400 / 1000 = 0.40 m/s2

 

Q2.3 If a body experiences a resultant force of 500 N and accelerates away at 2.0 m/s2, what is the mass of the body?

F = ma, so m = F / a = 500 / 2 = 250 kg

 

Q2.4 A bus of mass 2800 kg uniformly accelerates from 0 to 15 m/s in 20 seconds.

(a) Calculate the acceleration of the bus.

acceleration (m/s2) = change in speed (m/s) / time taken (s)

a = ∆v / ∆t = 15 / 20 = 0.75 m/s2

(b) Calculate the resultant force acting on the bus.

F = ma, substituting gives resultant force = 2800 x 0.75 = 2100 N

 

Q2.5 The engine of a van of mass 2000 kg generates a driving force of 6000 N.

If it experiences a drag force due to air resistance of 500 N, calculate its acceleration.

F = ma, a = F / m = 6000 / 2000 = 3 m/s2

 

Q2.6 What force is required to vertically accelerate a 8000 kg rocket at 3 m/s2? (g = 9.8 m/s2)

F1 = weight of rocket = downward force = 8000 x 9.8 = 78 400 N

F2 = upward force of the rocket to overcome gravity and accelerate vertically at 3 m/s2

F = ma, F = the net vertical force = F2 - F1, m = mass of rocket, a = acceleration of 3 m/s2

F2 - F1 = ma, F2 = F1 + ma

F2 = F1 + ma = 78 400 + (8000 x 3) = 102 400 N

 

Q2.7 If the force exerted on an object is quadrupled and the mass doubled, what happens to the acceleration?

F = ma, a = F/m (call it 1)

a = 4f / 2m = 2 F/m, so the acceleration is doubled.

 

Q2.8 High speed photography can be used to analyses the motion of sports equipment.

In cricket, a fast bowler sends the cricket ball down at 36 m/s (~80 mph, ~130 km/h) towards the batsman.

The cricket ball has a mass of 160 g and is a bit worn and slightly soft.

The batsman holds the bat still and blocks the ball which rebounds directly back towards the bowler at 30 m/s.

(a) If the contact time was found to be 0.002 seconds, what force did both the cricket ball experience and in what direction?

(i) You first need to calculate the acceleration - rate of change of velocity, and watch the signs, you are dealing with vector quantities!

a = ∆v / ∆t = (v - u) / ∆t, a = acceleration (m/s2), v and u = final and initial velocities, ∆t = time taken.

u = +36 m/s, v = -30 m/s, and ∆t = 0.002 s, a plus sign means the direction towards the batsman.

a = (-30) - (+36) / 0.002 = -33 000 m/s2

(ii) You can now calculate the force involved from Newton's 2nd Law equation.

F = ma,  F force in N, m = mass in kg, a = acceleration in m/s2

F = (160/1000) x 33 000 = - 5280 N

The cricket ball experiences a force of 5280 N back towards the bowler.

(b) A new ball is harder than a used ball and the contact time may only be 0.001 seconds.

What effect does this have on the force experienced by a new cricket ball compared to the worn ball in (a)?

This means ∆t is halved, the acceleration is doubled, so is the impact force is doubled, so the cricket ball experiences a force of 2 x 5280 = 10 560 N in the direction back towards the bowler.

(c) Why might the impact time be longer for the worn ball compared to a new ball?

The worn ball is a little softer and the change in velocity takes a bit longer - the softness has a cushioning effect that reduces the impact force.

 

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