If an object is moving at #4 m/s# over a surface with a kinetic friction coefficient of #u_k=80 /g#, how far will the object continue to move?

Question

If an object is moving at #4 m/s# over a surface with a kinetic friction coefficient of #u_k=80  /g#, how far will the object continue to move?

Julia Aguilar · Accepted Answer

The object will travel only #0.10m# (because a kinetic coefficient of #80/g# is very large!)

This problem is most easily done by conservation of energy. Two energy forms are involved: A change in kinetic energy: #1/2mv_f^2-1/2mv_i^2# Frictional heating: #muF_NDelta d# The equation for conservation of energy is, in this case #1/2mv_f^2-1/2mv_i^2+mumgDelta d=0# Notice that we can divide every term by the mass m, (including the right side of the equation) Also, #v_f=0# , and inserting the values for #mu#, #v_i# and g, we get: #-1/2(4)^2+(80/g)(g)Delta d=0# Cancel the two #g#'s #-1/2(4)^2+(80)Delta d=0# #-8+80Delatd=0# #80Deltad=8# #Deltad=0.1m#

Benjamin Asiedu · Answer

undefined To find the distance the object will continue to move, you can use the formula:

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

Where:
- $d$ is the distance traveled
- $v$ is the initial velocity of the object (4 m/s in this case)
- $\mu_k$ is the coefficient of kinetic friction (0.80 in this case)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s²)

Plugging in the values:

$$d = \frac{(4 \, \text{m/s})^2}{2 \times 0.80 \times 9.8 \, \text{m/s}^2}$$

$$d \approx 1.02 \, \text{m}$$

So, the object will continue to move approximately 1.02 meters.