Question

In a shuffleboard game, the puck slides on a horizontal floor before coming to complete stop. If the coefficient of kinetic friction between the puck and floor is 0.5 and the initial speed of the puck is 14.8m/s what distance did the puck travel (in meters before coming to a stop? Use kinematics and forces for in your solution (i.e. not energy).

Answer 1

22.33 m

Step-by-step explanation:

We are given;

Coefficient of kinetic friction between the puck and floor; μ = 0.5

initial speed of the puck; u = 14.8m/s

From Newton's equation of motion, we know that;

v² = u² + 2as

But since it decelerate to rest, the acceleration will be negative.

Thus;

v² = u² - 2as

Final velocity is zero, thus;

0 = u² - 2as

Thus, a = u²/2s

Where s is the distance covered before coming to rest.

Now, we know that formula for the frictional force is;

F = μmg

F/m = μg

We also know that F/m = a

Thus, a = μg

Thus:

u²/2s = μg

s = u²/(2μg)

s = 14.8²/(2 × 0.5 × 9.81)

s = 22.33 m