State Newton's laws of motion - Leaving Cert Physics - Question 6 - 2009

To demonstrate that F = ma is a special case of Newton's second law, we start with the definition of force as the rate of change of momentum:

$F = \frac{dp}{dt}$

Where momentum (p) is defined as:

$p = mv$

Taking the derivative:

$\frac{dp}{dt} = \frac{d(mv)}{dt}$

Assuming mass (m) is constant, this simplifies to:

$F = m \frac{dv}{dt}$

Since acceleration (a) is defined as:

$a = \frac{dv}{dt}$

We can conclude:

$F = ma$

Thus, F = ma is indeed a special case of Newton’s second law.

To calculate the average acceleration (a) of the skateboarder, we use the formula:

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

Where:

• Final velocity (v) = 12.2 m/s
• Initial velocity (u) = 0 (starts from rest)
• Distance (s) = 25 m

Substituting the values:

$12.2^2 = 0 + 2a(25)$

This simplifies to:

$148.84 = 50a$

Thus,

$a = \frac{148.84}{50} = 2.9768\ m/s^2 \approx 2.98\ m/s^2$

The weight of the skateboarder (W) is given by:

$W = mg$

Where:

• m = 70 kg (mass of skateboarder)
• g = 9.8 m/s² (acceleration due to gravity)

Calculating the weight:

$W = 70\ kg \times 9.8\ m/s^2 = 686\ N$

The component of the weight that acts parallel to the ramp is:

$W_{parallel} = W \sin(\theta)$

Substituting values:

$W_{parallel} = 686\ N \times \sin(20°) \approx 686 \times 0.342 = 234.41\ N$

The net force acting on the skateboarder on the ramp can be found using Newton's second law:

$F_{net} = ma = 70\ kg \times 2.98\ m/s^2 = 208.6\ N$

The force of friction (F_f) is found by:

$F_f = W_{parallel} - F_{net}$

Substituting the values:

$F_f = 234.41\ N - 208.6\ N = 25.81\ N$

The centripetal force (F_c) required to keep the skateboarder moving in a circle is given by:

$F_c = \frac{mv^2}{r}$

Where:

• m = 70 kg
• v = 10.5 m/s (speed)
• r = 10 m (radius)

Substituting these values:

$F_c = \frac{70\ kg \times (10.5)^2}{10} = \frac{70 \times 110.25}{10} = 771.75\ N$

To find the maximum height (h), we use the conservation of energy principle, where kinetic energy (KE) is converted into potential energy (PE):

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

Therefore,

$\frac{1}{2}(70)\ times (10.5)^2 = 70gh$

Cancelling mass (m) gives:

$\frac{1}{2}v^2 = gh$

Substituting in values:

$\frac{1}{2}(10.5)^2 = 9.8h$

Thus:

$\frac{110.25}{19.6} = h \approx 5.63 \ m$

The velocity-time graph of the skateboarder's motion would show a steady increase in velocity as he accelerates down the ramp, reaching a peak velocity of 12.2 m/s at the bottom. After that, the velocity remains constant at 10.5 m/s as he transitions into the circular ramp.