Question

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.

Answer 1

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  $v = 11.76 \ m/ s$

Explanation:

From the question we are told that

   The  total distance traveled is  $d = 1.2 \ m$

    The mass of the block is  $m_b = 0.3 \ kg$

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   $PE = KE$

Where  PE  is the potential energy which is mathematically represented as

      $PE = m * g * h$

substituting values

     $PE = 3 * 9.8 * 0.60$

      $PE = 17.64 \ J$

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          $KE = \frac{1}{2} * m v^2$

So

      $\frac{1}{2} * m* v ^2 = PE$

substituting values  

  =>    $\frac{1}{2} * 3 * v ^2 = 17.64$

=>       $v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }$

=>    $v = 11.76 \ m/ s$