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erastovalidia [21]
3 years ago
13

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.

Physics
1 answer:
kotegsom [21]3 years ago
4 0

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

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