Question

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

Answer 1

K_b = 78 J

Step-by-step 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