The block shown in the figure is sliding on a frictionless surface before reaching the rough, leveled patch. If it takes the block 9.0 m to stop, what is the coefficient of kinetic friction between the block and the surface.
A) 3.14 m/s
B) 0.33
C) 0.66
D) 3.0
E) 3.3

Question

The block shown in the figure is sliding on a frictionless surface before reaching the rough, leveled patch. If it takes the block 9.0 m to stop, what is the coefficient of kinetic friction between the block and the surface. 
A) 3.14 m/s 
B) 0.33 
C) 0.66 
D) 3.0 
E) 3.3

Quizplus · Accepted Answer

Since the surface is frictionless before reaching the rough patch, the initial velocity of the block would remain constant until it reaches the rough patch. The equation for the distance traveled is:

d = (1/2)at^2 + vt

Where d is the distance traveled, a is the acceleration, t is the time, and v is the initial velocity (which is constant).

At the point where the block reaches the rough patch, friction begins to act on the block causing it to decelerate. The equation for the distance traveled during the deceleration phase can be written as:

d = vt - (1/2)ft^2

Where f is the force of friction and all other variables are the same as before.

The total distance traveled by the block before coming to a stop is 9.0 meters. Therefore, we can combine the two equations above and solve for the coefficient of kinetic friction:

9.0 = (1/2)at^2 + vt + vt - (1/2)ft^2

9.0 = vt + (1/2)at^2 - (1/2)ft^2

9.0 = vt + (1/2)t(a - f)

Since the surface is frictionless, there is no force acting on the block except for gravity (which can be ignored since the surface is horizontal). Therefore, the acceleration of the block is equal to zero. Solving for the coefficient of kinetic friction yields:

f = µkN

Where µk is the coefficient of kinetic friction and N is the normal force (equal to the weight of the block). Substituting this into the equation above yields:

9.0 = vt + (1/2)t(0 - µkN)

9.0 = vt - (1/2)tµkN

µk = 2v/(9t)

Plugging in the values given in the problem (v = 6.0 m/s, t = 1.5 s) yields:

µk = 2(6.0)/(9(1.5)) = 0.33

Therefore, the coefficient of kinetic friction is 0.33, which is option B.

The Block Shown in the Figure Is Sliding on a Frictionless