Question

A cyclist is freewheeling (not exerting additional force) down a 7 degree angle hill. The cyclists weight is 75N. What acceleration is the cyclist experiencing? I have to do the following:1. Draw a free body diagram2. Identify Givens and Unknowns3. Identify the Equations4. Set up the equation using the givens and unknowns5. Solve

Answer 1

The free body diagram in shown below:

From the diagram and the problem we have that:

• The weight and angle of the inclined plane are given.

,

• The normal force and the components of the weight are unknown (this implies that the acceleration is unknown too); we also notice that the mass is not given the it is also an unknown.

We know that Newton's second law states that:

`\vec{F}=m\vec{a}`

where F is the resultant force and a is the acceleration. Since this is a vector equation we can decomposed it in two scalar equations (in this case we only need two scalar equations since the forces are coplanar), then we have:

`\begin{gathered} Wx=ma_x \\ N-W_y=ma_y \end{gathered}`

Since we don't expect the cyclist to move in the y direction (otherwise he will surely fall) the equations above would reduce to:

`\begin{gathered} W_x=ma \\ N-W_y=0 \end{gathered}`

From the first equation we can solve the acceleration, to do this we use the triangle to get the x-component of the weight:

`\begin{gathered} W_x=ma \\ W\sin \theta=ma \\ a=(W\sin \theta)/(m) \end{gathered}`

Since the weight is given but not the mass we use the fact that the weight is:

`W=mg`

to get the mass, then we have:

`\begin{gathered} m=(W)/(g) \\ m=(75)/(9.8) \\ m=7.65 \end{gathered}`

hence the mass of the cylcist is 7.65 kg.

Now that we have all the values we need we plug them in the expression for the acceleration:

`\begin{gathered} a=(75\sin 7)/(7.65) \\ a=1.19 \end{gathered}`

Therefore the acceleration of the cyclist is 1.19 meters per second per second.