If an object is moving at #4 ms^-1# over a surface with a kinetic friction coefficient of #u_k=3 /g#, how far will the object continue to move?

Question

If an object is moving at #4 ms^-1# over a surface with a kinetic friction coefficient of #u_k=3    /g#, how far will the object continue to move?

Amelia Adams · Accepted Answer

The distance is #=2.67m#

Let the mass of the object be #=mkg#

The acceleration due to gravity is #g=9.8ms^-2#

The initial velocity is #u=4ms^-1#

The final velocity is #v=0ms^-1#

The coefficient of kinetic friction is #mu_k=3/g#

The normal force is #N=mgN#

The frictional force is #=F_r N#

#mu_k=F_r/N#

#F_r=mu_k*N=3/g*mg=3mN#

According to Newton's Second Law

#F=ma#

#F_r=-ma#

The acceleration is #a=-F_r/m=-3m/m=-3ms^-2#

Apply the equation of motion,

#v^2=u^2+2as#

The distance is

#s=(v^2-u^2)/(2a)=(0-4^2)/(2*(-3))=16/6=8/3=2.67m#

Isabella Alvarez · Answer

undefined To find out how far the object will continue to move, we can use the equation for kinetic friction:

$f_k = \mu_k \times N$

Where:
- $f_k$ is the force of kinetic friction
- $\mu_k$ is the coefficient of kinetic friction
- $N$ is the normal force exerted on the object

Since the object is moving horizontally, the normal force ($N$) is equal to the gravitational force acting on the object:

$N = mg$

Where:
- $m$ is the mass of the object
- $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$)

Given that the coefficient of kinetic friction ($\mu_k$) is $3/g$, we can substitute the values into the equation:

$f_k = (3/g) \times mg$

$f_k = 3m$

The force of kinetic friction ($f_k$) is also equal to the mass of the object multiplied by its acceleration ($ma$). Since the object is moving at a constant velocity, the acceleration is $0$.

$f_k = ma$

$3m = 0$

Thus, the object will continue to move indefinitely because the force of kinetic friction is insufficient to stop it. Therefore, there is no specific distance it will travel over the surface.