Question

A motorcyclist is coasting with the engine off at a steady speed of 20.0 m/s but enters a sandy stretch where the coefficient of kinetic friction is 0.70. Will the cyclist emerge from the sandy stretch without having to start the engine if the sand lasts for 15 m? If so, what will be the speed upon emerging?

Short answer

The obtained value of speed upon emerging is $u=13.9\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Calculate the acceleration of the cyclist

In this problem, first, draw the free body diagram of the motorcyclist and consider that the cyclist is moving to the right.

The vertical acceleration is zero in this case; so the normal force will be equal to the body’s weight.

Given data:

The speed of the motorcycle is $v=20\mathrm{m}/\mathrm{s}$.

The coefficient of friction is ${\mu }_k=0.7$.

The distance to which sand lasts is $d=15\mathrm{m}$.

The free body diagram of the cyclist is as follows:

The relation to calculate the forces in the x-direction bucket is given by:

$\begin{array}{c}\Sigma F_x=ma\\ -F_f=ma\\ -{\mu }_{k}mg=ma\\ a=-{\mu }_{k}g\end{array}$

Here, $F_f$is the frictional force, a is the acceleration, m is the mass, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{l}a=-\left(0.7\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ a=-6.86\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 2. Calculate the distance covered by the cyclist

The relation to calculate the distance is given by:

$v{\text{'}}^2=v^2+2as$

Here, v’ is the final speed of the cyclist whose value is zero, and s is the distance covered by the cyclist.

On plugging the values in the above relation, you get:

$\begin{array}{c}{\left(0\right)}^2={\left(20\mathrm{m}/\mathrm{s}\right)}^2+2\left(-6.86\mathrm{m}/{\mathrm{s}}^2\right)s\\ s=29.1\mathrm{m}\end{array}$

Since the given distance of sand is $d=15\mathrm{m}$, it is clear that the cyclist emerges from the sand.

Step 3. Calculate the speed of the cyclist

The relation to calculate the speed is given by:

$u^2=v^2+2ad$

Here, u is the speed upon emerging from the sand.

On plugging the values in the above relation, you get:

$\begin{array}{c}{u}^2={\left(20\mathrm{m}/\mathrm{s}\right)}^2+2\left(-6.86\mathrm{m}/{\mathrm{s}}^2\right)\left(15\mathrm{m}\right)\\ u=13.9\mathrm{m}/\mathrm{s}\end{array}$

Thus, $u=13.9\mathrm{m}/\mathrm{s}$is the required speed of the cyclist.