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

To show that $F = ma$ is a special case of Newton's second law, consider:

$F = m \frac{dv}{dt} = m \frac{d}{dt}(v)$

If the mass (m) is constant, we can express the force as being equal to the mass times acceleration:

$F = ma$

This indicates that when the mass is constant, any net force acting on an object results in acceleration given by $a = \frac{F}{m}$.

Using the kinematic equation:

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

Here, $v = 12.2 \, \text{m/s}, u = 0, s = 25 \, \text{m}$, we get:

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

Solving for $a$ gives:

$a = \frac{12.2^2}{2 \times 25} = 2.977 \, \text{m/s}^2$

The weight of the skateboarder is given by:

$W = mg = 70 \times 9.8 = 686 \, \text{N}$

The component parallel to the ramp is:

$W_{\parallel} = W \sin(\theta) = 686 \sin(20°) = 234.63 \, \text{N}$

The net force acting on the skateboarder can be expressed as:

$F_{\text{net}} = mg \sin(20°) - F_{f}$

Thus,

$F_{f} = 234.63 - (70 \times 2.977) = 226.25 \, \text{N}$

The formula for centripetal force is given by:

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

Substituting the known values:

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

The potential energy at maximum height is equal to the kinetic energy when leaving the ramp:

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

Thus,

$h = \frac{v^2}{2g} = \frac{(10.5)^2}{2 \times 9.8} = 5.63 \, \text{m}$

The velocity-time graph would start at (0,0) and have a curve that rises steadily to the velocity at the bottom of the ramp (12.2 m/s), then levels off as it maintains the speed of 10.5 m/s until reaching the circular ramp.