Question

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Use work and energy to find how far the box slidesacross the rough surface before stopping.

Answer 1

Explanation:

The given data is as follows.

        k = 130 N/m,       $\Delta x$ = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

   $\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

          $v^{2} = \frac{3.757}{2.8}$

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = $\mu \times mg$

                                    = $0.15 \times 2.8 \times 9.8$

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           $\frac{1}{2}mv^{2} = F_{f} \times d$

           $\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

             1.8515 = $4.116 \times d$

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.