# Newton's Laws : Worked Examples

1. A car of mass 1000 kg travelling at 36 km/h is brought to rest over a distance of 20 m. Find:
(a) the average retardation; (b) the average braking force in newtons.

The retardation is found from the formula,
v² = u² + 2ax ( where v is the final velocity; u is the initial velocity; a is the acceletation and x is the distance)
so, v = 0 m/s  , u = 36 km/h = 36 × 1000 / 60 × 60 = 10 m/s
x = 20 m

By substitution,
0² = 10² + 2 × a × 20

therefore     a = - 10² / 2 × 20 = -2.5 m/s² ( minus sign means retardation) .

Knowing m and having found a, we now substitute in
F = ma,
to find F. Thus,
F= 1000 × 2.5 = 2500 N (
that is the average braking force )

P.s. If you want to know how I typed the multiplication sign (× ) and the superscript ( ² ), that is easy : just hold Alt 0215 to type × and hold Alt 0178 to type ².

2.  A bullet of mass 20 g, travelling with a velocity of 16 m/s, penetrates a sandbag and is brought to rest in 0.05 s.
Find : (a) the depth of penetration in metres;(b) the average retarding force of the sand in newtons.

We must first find the average retardation from the formula,
v = u + at (
where v=0, u = 16m /s  and t =0.05s)
Substituting,
0 = 16 + a × 0.05
from which      a = - 16 / 0.05 = - 320 m/s² ( minus sign means retardation) .

The depth of penetration may be found using either
x = ut + 1/2  at²
or
v² = u² + 2ax

Suppose we choose the latter, then
0² = 16² + 2(-320) × x
whence             x   = -16 × 16 / 2(-320) = 0.4 m

The average retardation force may now be calculated from
F = ma
in which         m = 20g = 0.02 kg ( we must convert to kg)and                 a  = -320 m/s²
Thus,                F = 0.02 × -320 = - 6.4 N ( minus sign means retardating force).

3. The valve of a cylinder containing 12 kg of compressed gas is opened and the cylinder empties in 1 min 30 s. If the gas issues from the exit nozzle with an average velocity of 25 m /s,
find the force exerted on the cylinder.

The force required to accelerate the gas out of the cylinder is given by,
F = ma
= mass ×  change in velocity/ time taken
= change in momentum/ time taken
The velocity of the gas changes from rest to 25 m / s
change in momentum = 12 × 25 kg m/s
time taken                        = 90 s
average force on gas     = 12 × 25/90 = 3.3 N

By Newton's third law, an equal reaction force is exerted on the cylinder;

average force on gas = average force on the cylinder = 3.3 N.

4. A 50g load is placed on a straight air-track sloaping at an angle of 45° to the horizontal. Calculate, in cm/s², the acceleration of the load as it is slides down and also the distance it would move from rest in 0.2 s. (
Assume negligible friction and take g = 980 cm/s².) This example illustrates the use of CGS units.

The force on the load causing it to accelerate is the resolved part of its weight along the air-track and is equal to
50 cos 45° gf.
Now, cos 45° = 0.707 and hence,
the accelerating force = 50 × 0.707 × 980 dyn

The acceleration is found from the equation F = mathus, a = F / m = 693 cm/s²

The distance, x, moved from rest is found from the equation    x = ut + 1/2 at²
where u = 0 cm/s
a= 693 cm/s²
t = 0.2 s
Substituting,
x = 0 × 0.2 + 693 × 0.2² /2
= 13.9 cm.

Now, my dear friend, try to solve this problem:
A man whose mass is 70 kg stands on a spring weighing machine inside an elevator. When the elevator starts to ascend its acceleration is 2.45 m / s². What is reading on the scales? What will the weighing machine read : (a) when the velocity of the elevator is uniform;(b) as it comes to rest with a retardation of 4.9 m/s²? ( Hint: revise  post

### Newton's second law of motion