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:

• The momentum is defined as $p = mv$, where $m$ is mass and $v$ is velocity.

• Therefore, if we take the derivative of momentum:

  F = rac{dp}{dt} = rac{d(mv)}{dt}

• Assuming mass is constant, this simplifies to:

  F = mrac{dv}{dt} = ma

This demonstrates that when mass is constant, Newton's second law leads to the form $F = ma$.

To find the average acceleration, use the equation of motion:

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

Where:

• $v = 12.2 \, ext{m/s}$ (final velocity)

• $u = 0 \,$ (initial velocity)

• $s = 25 \, ext{m}$ (distance)

• Rearranging:

  $a = \frac{v^2 - u^2}{2s} = \frac{(12.2)^2 - 0}{2 \times 25} = 2.977 \, \text{m/s}^2$

To find the component of the skateboarder's weight parallel to the ramp, use:

$W = mg \sin(\theta)$

Where:

• $m = 70 \, \text{kg}$ (mass)

• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)

• $\theta = 20°$ (angle)

• Thus:

  $W = 70 \times 9.8 \times \sin(20°) \approx 23.63 \, \text{N}$

The net force acting on the skateboarder is calculated by considering the weight parallel to the ramp and the mass's downwards force:

Let:

• $F_r = W - ma$

• Using the previously calculated values:

  $F_r = 23.63 - (70 \times 2.977) = 23.63 - 208.38 \approx -184.75 \, \text{N}$ The negative sign indicates that the force of friction acts upwards along the ramp.

The inward centripetal force can be calculated using:

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

Where:

• $v = 10.5 \, \text{m/s}$ (velocity)

• $r = 10 \, ext{m}$ (radius)

• Thus:

  $F_c = \frac{70 \times (10.5)^2}{10} = 77.175 \, ext{N}$

To find the maximum height, use the conservation of energy principle:

The mechanical energy at the bottom will equal the potential energy at the maximum height:

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

Thus, solving for height $h$ gives:

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

The velocity-time graph should start from the origin (0,0), showing an increasing linear slope initially as the skateboarder accelerates down the ramp, then a level section representing constant velocity as the skateboarder goes horizontally, and finally, a steep incline if the graph indicates going downwards on the circular ramp, eventually reaching a peak at the maximum height.