If a #1 kg# object moving at #14 m/s# slows down to a halt after moving #7 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

Question

Adrian Ayala · Accepted Answer

#W_"net" = -3/8(14)^2 J = mu_k (mg)*d ~~ 70mu_k #
Solve for #mu_k ~~ 21/20# This seem too excessive of a coefficient of kinetic friction... See below for possible explanation...

We can use the work energy theorem, which states that the net work done is equal to the change in Kinetic Energy:
#W_"net" = DeltaKE = 1/2 m(v_f^2 - v_i^2) = 1/2 m((1/2v_i)^2 - v_i^2)#
#W_"net" = 1/2 m(-3/4v_i^2)= -3/8mv_i^2#;
Now with #m = 1; v_i = 14 m/s#
#W_"net" = -3/8(14)^2 J#

Now let's draw a Free Body Diagram:
From FBD we see that the horizontal force slowing the object is the force of friction acting in the opposite direction of motion. And knowing that #W = F*d #
#F_f = mu_k (mg)# this the force doing work hence:
#W_"net" = -3/8(14)^2 J = mu_k (mg)*d ~~ 70mu_k #
Solve for #mu_k ~~ 21/20# Now this amount of kinetic friction is
excessive. What went wrong? Well the general approach is correct but the assumption that friction is the only one consuming energy in slowing the object may not be correct. One can argue Heat and Sound will also take energy away lowering the energy available for friction and hence approaching a reasonable answer of #mu_k < .4#

Benjamin Asiedu · Answer

undefined The coefficient of kinetic friction can be calculated using the formula:

$$ \text{Coefficient of Kinetic Friction} = \frac{\text{Force of Kinetic Friction}}{\text{Normal Force}} $$

The force of kinetic friction $ F_k $ can be calculated using Newton's second law:

$$ F_k = \mu_k \times N $$

Where $ \mu_k $ is the coefficient of kinetic friction and $ N $ is the normal force.

Given that the object comes to a halt, we know that the force of kinetic friction is equal to the force applied:

$$ F_k = m \times a $$

The acceleration $ a $ can be calculated using the equation of motion:

$$ v^2 = u^2 + 2as $$

where $ u $ is the initial velocity, $ v $ is the final velocity, $ s $ is the displacement, and $ a $ is the acceleration.

Rearranging the equation to solve for acceleration $ a $:

$$ a = \frac{v^2 - u^2}{2s} $$

Substituting the given values:

$$ a = \frac{(0)^2 - (14 \, \text{m/s})^2}{2 \times 7 \, \text{m}} $$

$$ a = \frac{-196}{14} \, \text{m/s}^2 $$

$$ a = -14 \, \text{m/s}^2 $$

Now, substituting the calculated acceleration into the equation for the force of kinetic friction:

$$ F_k = m \times a $$

$$ F_k = (1 \, \text{kg}) \times (-14 \, \text{m/s}^2) $$

$$ F_k = -14 \, \text{N} $$

Since the object comes to a halt, the force of kinetic friction is equal in magnitude to the force applied. Therefore:

$$ \text{Force of Kinetic Friction} = 14 \, \text{N} $$

The normal force $ N $ can be calculated using Newton's second law:

$$ N = mg $$

$$ N = (1 \, \text{kg}) \times (9.8 \, \text{m/s}^2) $$

$$ N = 9.8 \, \text{N} $$

Now, substituting the calculated values into the formula for the coefficient of kinetic friction:

$$ \text{Coefficient of Kinetic Friction} = \frac{\text{Force of Kinetic Friction}}{\text{Normal Force}} $$

$$ \text{Coefficient of Kinetic Friction} = \frac{14 \, \text{N}}{9.8 \, \text{N}} $$

$$ \text{Coefficient of Kinetic Friction} = 1.43 $$

Therefore, the coefficient of kinetic friction of the surface is approximately $ 1.43 $.