Question

A body sliding on ice with a velocity of 8m/s comes to rest after travelling 40m.find the coefficient of friction

Answer 1

Let,u be initial velocity and v be final velocity.
s be displacement and a be the acceleration.
K be the coefficient of friction.
From kinematics,
v² = u² + 2as
0 = 8² - 2a(40) (since, it's decelerating)
a= 0.8
a = K.g = 0.8 (g is acceleration due to gravity, value = 10m/s²)
K = 0.08

Answer 2

The answer to the question will be 0.082 .

CALCULATION:

As per the question, the initial velocity of the body u = 8 m/s.

The final velocity of the body v = 0 m/s.

The stopping distance covered by the body s = 40 m.

From equation of kinematics, we know that -

                                       $v^2-u^2=\ 2as$

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

                                      ⇒ $a=\frac{0^2-8^2}{2\times 40}$

                                             $=\frac{-64}{80}\ m/s^2$

                                             $=\ -0.8\ m/s$

The negative sign is used due to the fact that velocity is decreasing due to the opposing force i.e friction.

Let us consider the mass of the body is m and $\mu$ is the coefficient of friction.

The frictional force is calculated as F = $\mu mg$   [1]

Here, g is the acceleration due to gravity.

From Newton's second law of motion we know that net force is the product of mass with acceleration.

Mathematically F = ma.     [2]

From 1 and 2 we get , $\mu mg=\ ma$

                                          ⇒ $\mu g=\ a$

                                          ⇒ $\mu=\ \frac{a}{g}$

                                                  $=\ \frac{0.8}{9.8}$

                                                  $=\ 0.082$            [ans]