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shepuryov [24]
3 years ago
10

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

g it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.9 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 16.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

The velocity is v_b = 20.17 \ m/s

Explanation:

From the question we are told that

   The mass of the ball is  m = 0.245 \ kg

   The radius is  r =  59.8 \  cm  =  0.598 \ m

   The force is  F =  30.9 \ N

   The speed of the ball is  v = 16.0 \ m/s.

Generally the kinetic energy at the top of the circle is mathematically represented as

    K_t  =  \frac{1}{2} *  m  *  v^2

=> K_t  =  \frac{1}{2} *  0.245   *  16.0 ^2  

=> K_t  =  31.36 \ J  

Generally the work done by the force applied on the ball from the top to the bottom  is mathematically represented as

       W =  F *  d

Here  d is the length of  a semi - circular arc which is mathematically represented as

       d =  \pi * r

So

      W =  30.9 *  0.598

      W = 18.48 \ J

Generally the kinetic energy at the bottom is mathematically represented as

      K_b  =  \frac{1}{2} *  m *  v_b^2

=>    K_b  =  \frac{1}{2} *  0.245  *  v_b^2

=>   K_b  =  0.1225  *  v_b^2

From the law of energy conservation

     K_t +  W  =K_b

=>    31.36+  18.48 = 0.1225  *  v_b^2

=>    v_b = 20.17 \ m/s

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