Question

Water falls without splashing at a rate of 0.400 L/s from a height of 3.60 m into a 0.530-kg bucket on a scale. If the bucket is originally empty, what does the scale read 3.40 s after water starts to accumulate in it?

Answer 1

using conservation of energy,

kinetic energy of water = potential energy of water

1/2 m_w v_w² = m_w g h

v_w = 2 g h                                                    { eq.1 }

where, g = 9.8 m/s²

h = water falls from height = 3.60 m

inserting these values in eq.1,

v_w = 2 (9.8 m/s²) (3.60 m)

v_w = 66.64 m²/s²

v_w = 8.4 m/s

Force due to momentum of water being stopped by the bucket will be given as :

F_m = m_w v_w                                       { eq.2 }

where, m_w = mass of water= 0.400 L/s = 0.400 kg/s

inserting the values in eq.2,

F_m = (0.400 kg/s) (8.4 m/s)

F_m = 3.36 N

net weight of the bucket, W_net = weight of the bucket + weight of water @ 3.40 sec

W_net = m_b g + m_w g (3.40 sec)                                     

W_net = g [m_b + m_w (3.40 sec)]    { eq.3 }

where, m_b = mass of the bucket = 0.530 kg

inserting the values in eq.3,

W_net = (9.8 m/s²) [(0.530 kg) + (0.400 kg/s) (3.40 sec)]

W_net = (9.8 m/s²) (1.89 kg)

W_net = 18.522 N

The scale reading will be given as,   R = F_m + W_net                                        { eq.4 }

inserting the values in eq.4,

R = (3.36 N) + (18.522 N)

R = 21.882 N