Question

An object of mass m = 2 kg placed * on a rough horizontal surface is given an initial speed of 10 m / s and comes to rest after covering a distance d = 12m . The kinetic friction f_k is equal to :

Answer 1

We are given that a 20 kg object moves a distance of 12 meters. The intial velocity is 10 m/s. To calculate the friction force we will use the fact that the chage in the kinetic energy is equal to the work done by the frictio force, therefore, we have:

$\frac{1}{2}m(v_f^2-v_0^2)=W_f$

Where:

$\begin{gathered} m=\text{ mass} \\ v_f,v_0=\text{ final and initial velocities} \\ W_f=\text{ work done by friction} \end{gathered}$

Now, we plug in the values:

$\frac{1}{2}(2kg)((10\frac{m}{s})^2-0)=W_f$

Solving the operations:

$100J=W_f$

Now, the work done by friction is equal to the friction force multiplied by the distance:

$100J=F_fd$

Now, we divide both sides by the distance:

$\frac{100J}{12m}=F_f$

Solving the operations:

$8.33N=F_f$

Therefore, the force of friction is 8.33 Newtons.