Question

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point circled A, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vA = 13.4 m/s, and the block experiences an average frictional force of 7.00 N while sliding up the track. What is x? and what is the speed of the block if it reaches the top?

Answer 1

Answer:

The value of x is  $x = 0.1955\ m$

The value of the velocity at the top is  $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

     The mass of the block is $m_b = 0.490kg$

      The distance of compression is $x$

      The force constant is $k = 450 N/m$

      The radius of the circular track is $R =1.00m$

       The speed at the bottom of the track is $v_A = 13.4 \ m/s$

        The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy  of the block . So,

          $PE \ of \ spring = KE \ of \ block$

mathematically i.e

        $\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

       $0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

           $x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

              $= 0.1955\ m$

The average workdone by friction is  $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

                                         $W_f = 7 * 3.142$

                                               $= 21.99 J$

The kinetic energy at the bottom is

                                     $KE = \frac{1}{2} * 0.490 * 13.4^2$

                                            $= 43.99 \ N$

The potential energy gained at the top of the circle is

                              $PE = m_bgh$

Here h  is the height which is equal to d(diameter)  = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

                       $PE = 0.490 * 9.8 *2$

                             $=9.604 J$

Since energy is can not be created nor destroyed but transformed  according to first law of thermodynamics

                  $KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

                      $43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

                        $12.369= [2.45* v_{top} ^2]$

                            $v_{top} =\sqrt{\frac{12.369}{2.45} }$

                                  $= 2.25 \ m/s$