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

This page contains online questions only. Jot down
your answers and check them against the worked out answers at the end of
the page

**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/s^{2})

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/s^{2}?

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/s^{2},
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/s^{2}?
(g = 9.8 m/s^{2})

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.

SITEMAP
Website content © Dr
Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries and references to GCSE science course specifications
are unofficial.**
Using SEARCH some initial results may be ad links you
can ignore - look for docbrown**

INDEX for physics notes on
Newton's Laws of Motion