If an object is moving at #8 m/s# over a surface with a kinetic friction coefficient of #u_k=9 /g#, how far will the object continue to move?

Question

If an object is moving at #8 m/s# over a surface with a kinetic friction coefficient of #u_k=9    /g#, how far will the object continue to move?

Adrian Ayala · Accepted Answer

The distance is #=3.56m#

The coeffcient of kinetic friction is #mu_k=9/g#

#mu_k=F_r/N#

Let the mass of the object be #=m kg#

Then,

The normal reaction is #N=mg N#

The frictional force is

#F_r=mu_kN=mu_kmg=9/g*mg=9mN#

According to Newton's Second Law

#F=ma#

The acceleration is #a=F/m=F_r/m=-9m/m=-9ms^-2#

The initial velocity is #u=8ms^-1#

Apply the equation of motion

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

The final velocity #v=0#

#0=64-2xx9xxs#

The distance is

#s=64/18=3.56m#

Nathan Ashcroft · Answer

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

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

Where:
- $ d $ is the distance traveled,
- $ v $ is the initial velocity of the object (8 m/s in this case),
- $ \mu_k $ is the kinetic friction coefficient (9/g in this case),
- $ g $ is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values:

$$ d = \frac{{8^2}}{{2 \cdot 9/g \cdot 9.8}} $$

$$ d ≈ \frac{{64}}{{2 \cdot (9/9.8) \cdot 9.8}} $$

$$ d ≈ \frac{{64}}{{2 \cdot (0.918) \cdot 9.8}} $$

$$ d ≈ \frac{{64}}{{2 \cdot 0.918 \cdot 9.8}} $$

$$ d ≈ \frac{{64}}{{17.964}} $$

$$ d ≈ 3.56 \text{ meters} $$