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andriy [413]
3 years ago
7

Help!!

Physics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

The velocity at B is  v = 12.6 \ m/s  

The velocity at C is  v_c =10.80 \ m/s

Explanation:

From the question we are told that

     The mass of the bead is m_b  = 0.368 \ kg

     The first height is  h_1 = 4.20 \ m

      The second height is  h_2 = 2.14 \ m

      The initial speed of the bead is  u  = 1.96 \ m/s

According to the law of energy conservation

       KE _ A + PE_A =  KE_B + PE_B

Now KE _ i is the kinetic energy at A and it is mathematically represented as

        KE_A  =  \frac{1}{2} mu^2

         PE_i is the potential energy at A which is mathematically represented as

       PE_A =  mg h_1

       KE_f is the kinetic energy at B which is mathematically represented as

           KE_B =  \frac{1}{2} mv^2

Where v is the speed of the bead at B

           PE_B is the potential energy at B which equal to 0 because height is  0 at B

So

      \frac{1}{2} mu^2 + mg h_1 =  \frac{1}{2} mv^2

Making v the subject

       v = \sqrt{u ^2 + 2gh_1}

substituting values

     v = \sqrt{1.96 ^2 + 2* 9.8 * 4.20}

      v = 12.6 \ m/s  

According to the law of energy conservation

       KE _ B + PE_B =  KE_C + PE_C

So

      KE_B = \frac{1}{2} m v^2

      PE_B  = 0

KE_C is the kinetic energy at c which is mathematically represented as

         KE_C  = \frac{1}{2}  *   m v_c^2

 PE_C is the potential energy at C which is mathematically represented as

        PE_C  = mg h_2

So

      \frac{1}{2} m v^2  =   \frac{1}{2}  *   m v_c^2 +  mg h_2

     v^2  =    v_c^2 +  2g h_2

making v_c the subject

     v_c = \sqrt{12.6^2 - 2* 9.8 * 2.14}

      v_c =10.80 \ m/s

   

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