Question

In the Soapbox Derby in (Figure 1) , young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. Assume that the track begins with a 49-ft-long(1 m = 3.28 ft) section tilted 14? below horizontal.

What is the maximum possible acceleration of a car moving down this stretch of track?

Express your answer to two significant figures and include the appropriate units.

If a car starts from rest and undergoes this acceleration for the full l, what is its final speed in m/s?

Express your answer using two significant figures.

Answer 1

Concepts and reason

The concept used to solve this problem is force and final velocity equation.

The second law of Newton states that the net force on an object is the product of the mass of the object and the acceleration.

The force due to gravity acting on the object which is rolling down a hill can be resolved in the vertical and the horizontal components.

The final speed of the car can be calculated using the kinematics equation.

Fundamentals

The force is,

$F = ma$

Here, F is the force, m is the mass, and a is the acceleration.

The acceleration can be calculated using the Newton’s Second law.

The force due to gravity is,

${F_{\rm{g}}} = mg$

Here, ${F_{\rm{g}}}$ is the force due to gravity, m is the mass of the object, and g is the acceleration due to gravity.

The kinematics equation for the speed of the car is,

${v^2} - {u^2} = 2as$

Here, $v$ is the final speed of the car, $u$ is the initial speed, $a$ is the acceleration, and $s$ is the distance.

The following diagram shows the car rolling down a hill:

Here, ${F_{\rm{g}}}$ is the force due to gravity, N is the normal force, ${f_{\rm{s}}}$ is the frictional force, ${F_{\rm{g}}}\cos \theta$ and ${F_{\rm{g}}}\sin \theta$ are the components of the force due to gravity. Since there is low friction between the road and the wheels, ignore the friction. The normal force will be balance by ${F_{\rm{g}}}\cos \theta$ , the component of force due to gravity.

Balance the forces.

The net force is,

${F_{{\rm{net}}}} = {F_{\rm{g}}}\sin \theta - {f_{\rm{s}}}$

Substitute mg for ${F_{\rm{g}}}$ , and 0 for ${f_{\rm{s}}}$ , and ma for ${F_{{\rm{net}}}}$ .

$ma = mg\sin \theta$

Rearrange the equation.

$a = g\sin \theta$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g and $14^\circ$ for $\theta$ .

$\begin{array}{c}\\a = \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\sin \left( {14^\circ } \right)\\\\ = 2.4{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The maximum possible acceleration of the car moving down the stretch of track is $2.4{\rm{ m/}}{{\rm{s}}^2}$ .

Calculate the final speed of the car.

The equation for velocity is,

${v^2} - {u^2} = 2as$

Substitute 0 m/s for u.

$\begin{array}{c}\\{v^2} - {\left( {0{\rm{ m/s}}} \right)^2} = 2as\\\\v = \sqrt {2as} \\\end{array}$

Substitute $2.37{\rm{ m/}}{{\rm{s}}^2}$ for a and 49 ft for s in the above equation.

$\begin{array}{c}\\v = \sqrt {2\left( {2.37{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {49{\rm{ ft}}} \right)\left( {\frac{{0.3048{\rm{ m}}}}{{1{\rm{ ft}}}}} \right)} \\\\ = 8.4{\rm{ m/s}}\\\end{array}$

The final speed of the car is 8.4 m/s.

Ans:

The maximum possible acceleration of the car moving down the stretch of track is $2.4{\rm{ m/}}{{\rm{s}}^2}$ .

The final speed of the car is 8.4 m/s.