Question

9) A cardboard box of unknown mass is sliding upon a horizontal frictionless
surface. The box has a velocity of 4.56 m/s when it encounters friction. After
sliding 0.700 m, the box has a velocity of 3.33 m/s. What is the coefficient of
friction of the surface?

Answer 1

The coefficient of friction of the surface is 0.707.

Explanation:

According to Work Energy theorem, the change in kinetic energy of the object is equal to the work done on the object. And the work done is measure of force required to displace an object. So work done is the product of force and displacement. Thus, change in kinetic energy will be equal to the product of force and displacement.

  $Change in kinetic energy = Work done = Force \times displacement$

As force acting on the box will comprise of frictional force so,

         $\text {Force}=\mu \times m \times g$

        $Change in kinetic energy =\mu \times m \times g \times s$

        $\frac{1}{2} \times m \times\left(u^{2}-v^{2}\right)=\mu \times m \times g \times s$

'm' presents on both sides, so gets cancelled. Hence, the equation would be

          $\frac{1}{2} \times\left(4.56^{2}-3.33^{2}\right)=\mu \times 9.8 \times 0.7$

        $\mu=\frac{\left(4.56^{2}-3.33^{2}\right)}{2 \times 9.8 \times 0.7}=\frac{20.79-11.0889}{13.72}=\frac{9.7011}{13.72}=0.707$

So, the coefficient of friction of the surface is 0.707.