An object, previously at rest, slides #9 m# down a ramp, with an incline of #(pi)/6 #, and then slides horizontally on the floor for another #24 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Question

Nora Arzate · Accepted Answer

#k~=0,142# #pi/6=30^o# #E_p=m*g*h " Potential Energy of Object"# #W_1=k*m*g*cos 30*9 # #"Lost energy because friction on inclined plane"# #E_p-W_1": energy when object on ground"# #E_p_W_1=m*g*h-k*m*g*cos 30^o*9# #W_2=k*m*g*24 " lost energy on the floor"# #k*cancel(m*g)*24=cancel(m*g)*h-k*cancel(m*g)*cos 30^o*9# #24*k=h-9*k*cos 30^o# #"using "cos 30^o=0,866 ;h=9*sin30=4,5 m# #24*k=4,5-9*k*0,866# #24*k+7,794*k=4,5# #31,794*k=4,5# #k=(4,5)/(31,794)# #k~=0,142#

Nicholas Agrusa · Answer

undefined To find the kinetic friction coefficient, we need to use the equations of motion and the principle of conservation of energy.

The object's potential energy on the ramp is converted into kinetic energy at the bottom of the ramp and then into kinetic energy as it slides horizontally.

Using the equation for gravitational potential energy and the fact that the object starts from rest:
$mgh = \frac{1}{2}mv^2$

The height $h$ can be found using trigonometry:
$h = 9 \sin(\pi/6)$

Then, we have:
$mgh = \frac{1}{2}mv^2$ for the ramp part
and
$F_{\text{friction}} \times 24 = \frac{1}{2}mv^2$ for the horizontal part

From the first equation:
$mgh = \frac{1}{2}mv^2$

From the second equation:
$F_{\text{friction}} = \frac{1}{2}mv^2 / 24$

Substituting $mgh$ from the first equation:
$F_{\text{friction}} = \frac{mgh}{24}$

Then, substitute $h = 9 \sin(\pi/6)$ into the equation and solve for the kinetic friction coefficient.