ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

thermal energy is the answer