If an object is moving at #50 m/s# over a surface with a kinetic friction coefficient of #u_k=8 /g#, how far will the object continue to move?

Question

If an object is moving at #50 m/s# over a surface with a kinetic friction coefficient of #u_k=8   /g#, how far will the object continue to move?

Layla Ahmed · Accepted Answer

I found #156m# Considering that the only force acting on the object will be kinetic friction #f_k#, it will produce an opposite acceleration slowing down the object. From Newton's Second Law (horizontally): #-f_k=ma# and so: #a=-f_k/m# Friction will be: #f_k=u_kN# where, in this case, the normal reaction, #N#, will be equal to the weight #W=mg#; so: #a=-(u_kcancel(m)g)/cancel(m)=-8/g*g=-8m/s^2# With this acceleration you can use: #color(red)(v_f^2=v_i^2+2ad)# with #v_f=0# (the object stops) so you get: #0^2=50^2-(2*8*d)# #d=2500/(16)=156m#

Julia Aguilar · Answer

undefined To calculate the distance the object will continue to move, we can use the equation:

$ d = \frac{v^2}{2 \mu_k g} $

Where:
- $ d $ is the distance
- $ v $ is the initial velocity of the object (50 m/s in this case)
- $ \mu_k $ is the coefficient of kinetic friction (given as $ \mu_k = \frac{8}{g} $)
- $ g $ is the acceleration due to gravity (approximately 9.8 m/s²)

Plugging in the values:

$ d = \frac{(50 \, \text{m/s})^2}{2 \times \frac{8}{9.8} \times 9.8 \, \text{m/s}^2} $

$ d = \frac{2500 \, \text{m}^2/\text{s}^2}{2 \times \frac{8}{9.8}} $

$ d = \frac{2500 \, \text{m}^2/\text{s}^2}{16/9.8} $

$ d = \frac{2500 \times 9.8}{16} $

$ d = \frac{24500}{16} $

$ d \approx 1531.25 \, \text{m} $

So, the object will continue to move approximately 1531.25 meters.