A motorcycle and 60.0 kg rider accelerate at$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$up a ramp inclined $\mathbf{10}\mathbf{°}$above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

1. m = 60kg
2. $a=3.0m/s^2$
   
3. $\theta =10°$
   

Newton’s second law states that the net force on the object is the product of mass and the acceleration of the object. We have been given the acceleration and mass of the object, using the equation for Newton’s second law of motion, we can find out the net force acting on the object. Using the concept of vector resolution, we can find the force exerted by the motorcycle on the rider.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{a}$

Here, ${\mathbf{F}}_{\mathbf{net}}$ is the force, is the mass of the object, and a is the acceleration of the object.

(a)

Net Force is calculated using Newton’s law of motion as,

$\begin{array}{l}{F}_{net}=m\times a\\ =60\mathrm{kg}\times 3m/s^2\\ =180N\end{array}$

Therefore, the net force is 180N

(b)

Free body diagram:

The x-components of mg is calculated as,

$\begin{array}{l}{F}_x=mg\times sin(10°)\\ =60\mathrm{kg}\times 9.81m/s^2\times sin(10°)\\ =102.105N\end{array}$

The y-components of mg is calculated as,

$\begin{array}{l}{F}_y=mg\times \cos (10°)\\ =60\mathrm{kg}\times 9.81m/s^2\times \cos (10°)\\ =579.66N\end{array}$

As there is acceleration in x direction,

$$\begin{gathered} F-F_x=F_{net} \\ F-101.105N=180N \\ F=282.105N \end{gathered}$$

So, Force on rider from motorcycle

$$\begin{gathered} F_r=\sqrt{282.{105}^2+579.{66}^2} \\ =644.66N \end{gathered}$$

The force on the rider from the motorcycle in two significant figures is 640 N.