Define (i) velocity, (ii) friction - Leaving Cert Physics - Question 6 - 2009

Friction is the force that opposes the relative motion of two surfaces in contact. It acts parallel to the surfaces and is dependent on the nature of the surfaces and the normal force pressing them together.

To calculate the time taken to reach top speed, we use the formula:

$t = \frac{v - u}{a}$ where:

• $v = 50$ m/s (final velocity)
• $u = 0$ m/s (initial velocity)
• $a = 0.5$ m/s² (acceleration)

Substituting the values:

$t = \frac{50 - 0}{0.5} = 100 \text{ seconds}$

During the time the train travels at its top speed, the distance can be calculated using:

$d = vt$ where:

• $v = 50$ m/s
• $t = 90 \text{ minutes} = 5400$ seconds

Substituting the values:

$d = 50 \times 5400 = 270000 \text{ m}$

To find the acceleration when the brakes are applied, we can use the following formula:

$v^2 = u^2 + 2as$ where:

• $v = 0$ m/s (final velocity)
• $u = 50$ m/s (initial velocity)
• $s = 500$ m (stopping distance)

Rearranging for $a$:

$0 = 50^2 + 2a(500)$
$0 = 2500 + 1000a$
$1000a = -2500$
$a = -2.5 \text{ m/s}²$

Using Newton's second law, $F = ma$:

• $m = 30000$ kg
• $a = -2.5$ m/s²

The force is:

$F = 30000 \times -2.5 = -75000 \text{ N}$ The negative sign indicates that the force is acting in the opposite direction of the train's motion.

The kinetic energy (KE) lost can be calculated using the formula:

$KE = \frac{1}{2}mv^2$ Substituting the values:

$KE = \frac{1}{2} \times 30000 \times 50^2$ $= \frac{1}{2} \times 30000 \times 2500 = 37500000 \text{ J}$

The kinetic energy lost by the train is transformed primarily into heat and sound energy due to friction between the train's brakes and the wheels as it stops.

Force A is the driving force generated by the engines of the train, while Force B is the frictional force opposing the motion of the train.

When the force A is equal to the force T, the train moves at a constant speed because the net force acting on it is zero, resulting in no acceleration.

In the velocity-time graph:

• The graph starts from the origin, rises linearly until reaching 50 m/s (top speed), remains constant for 90 minutes, and finally slopes down to the x-axis as the train comes to a stop in 500 m.

The graph depicts an increasing section followed by a horizontal line and then a decreasing section as the train decelerates.