If a #1 kg# object moving at #7 m/s# slows down to a halt after moving #140 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

Question

Sebastian Acosta · Accepted Answer

I found approx #0.02# We can start by using: #color(red)(v_f^2=v_i^2+2ad)# to get: #0=7^2+2a*140# so that: #a=-49/280=-0.175m/s^2# We then use Newton´s Second Law #SigmavecF=mveca# along the #x# direction only and with only kinetic friction #f_k# acting to stop the object: #-f_k=ma# #-mu_k*N=ma# where the normal reaction is #N="weight"=mg# so: #-mu_k*1*9.8=1*(-0.175)# so: #mu_k=0.175/9.8=0.0178~~0.02#

Adam Arnold · Answer

undefined To find the coefficient of kinetic friction, we can use the equation: $ F_k = \mu_k \times N $, where $ F_k $ is the force of kinetic friction, $ \mu_k $ is the coefficient of kinetic friction, and $ N $ is the normal force.

First, we need to calculate the force of kinetic friction using the equation: $ F_k = m \times a $, where $ m $ is the mass of the object and $ a $ is the acceleration.

Given:
$ m = 1 \, \text{kg} $
$ v_i = 7 \, \text{m/s} $ (initial velocity)
$ v_f = 0 \, \text{m/s} $ (final velocity)
$ d = 140 \, \text{m} $ (distance traveled)

Using the equation $ v_f^2 = v_i^2 + 2ad $, we can solve for the acceleration ($ a $).

$ 0 = (7)^2 + 2 \times a \times 140 $
$ a = \frac{{(-7)^2}}{{2 \times 140}} $
$ a = -\frac{{49}}{{280}} \, \text{m/s}^2 $

Now, we can calculate the force of kinetic friction using $ F_k = m \times a $.

$ F_k = 1 \times \left(-\frac{{49}}{{280}}\right) $
$ F_k = -\frac{{49}}{{280}} \, \text{N} $

Since the object comes to a halt, the force of kinetic friction is equal in magnitude to the force applied (since there is no net force). Therefore, $ F_k = -\mu_k \times N $, where $ N $ is the normal force.

Now, we need to find the normal force. Since the object is moving horizontally, the normal force is equal to the gravitational force acting on the object, which is $ m \times g $.

$ N = m \times g $
$ N = 1 \times 9.8 $
$ N = 9.8 \, \text{N} $

Now, we can find the coefficient of kinetic friction ($ \mu_k $).

$ -\frac{{49}}{{280}} = -\mu_k \times 9.8 $
$ \mu_k = \frac{{49}}{{280 \times 9.8}} $
$ \mu_k \approx 0.177 $

Therefore, the coefficient of kinetic friction of the surface is approximately $ 0.177 $.