If a #5 kg# object moving at #6 m/s# slows to a halt after moving #2 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

Question

Nathan Ashcroft · Accepted Answer

The coefficient of kinetic friction is #=0.92#

The mass of the object is #m=5kg#

The initial speed of the object is #u=6ms^-1#

The final speed of the object is #v=0ms^-1#

The distance is #s=2m#

Apply the equation of motion

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

to calculate the acceleration

The acceleration is

#a=(v^2-u^2)/(2s)=(0^2-6^2)/(2*2)=-36/4=-9ms^-2#

According to Newton's Second Law

#F=ma#

The frictional force is #F_r=5*9=45N#

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

The normal force is #N=mg=5*9.8=49N#

The coefficient of kinetic friction is

#mu_k=F_r/N=45/49=0.92#

Leo Abeyta · Answer

undefined To find the coefficient of kinetic friction, we can use the equation:

$$ F_{\text{friction}} = \mu_k \times F_{\text{normal}} $$

Where:
- $ F_{\text{friction}} $ is the force of friction,
- $ \mu_k $ is the coefficient of kinetic friction,
- $ F_{\text{normal}} $ is the normal force.

First, we need to find the force of friction using Newton's second law:

$$ F_{\text{friction}} = m \times a $$

Where:
- $ m $ is the mass of the object (5 kg),
- $ a $ is the acceleration.

We know the object slows down to a halt, so the final velocity is 0 m/s. Using the equation of motion:

$$ v^2 = u^2 + 2as $$

Where:
- $ v $ is the final velocity (0 m/s),
- $ u $ is the initial velocity (6 m/s),
- $ a $ is the acceleration,
- $ s $ is the displacement (2 m).

Rearranging the equation to solve for acceleration:

$$ a = \frac{{v^2 - u^2}}{{2s}} $$

$$ a = \frac{{0^2 - 6^2}}{{2 \times 2}} = -9 \, \text{m/s}^2 $$

Now, using Newton's second law:

$$ F_{\text{friction}} = m \times a = 5 \, \text{kg} \times (-9 \, \text{m/s}^2) = -45 \, \text{N} $$

Since the object is slowing down, the force of friction acts in the direction opposite to the motion, hence the negative sign.

The normal force can be calculated as:

$$ F_{\text{normal}} = m \times g $$

Where:
- $ g $ is the acceleration due to gravity (9.8 m/s²).

$$ F_{\text{normal}} = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} $$

Now, we can find the coefficient of kinetic friction:

$$ \mu_k = \frac{{F_{\text{friction}}}}{{F_{\text{normal}}}} $$

$$ \mu_k = \frac{{-45 \, \text{N}}}{{49 \, \text{N}}} $$

$$ \mu_k \approx -0.92 $$

However, coefficients of friction cannot be negative, so we take the absolute value:

$$ \mu_k \approx 0.92 $$