A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14 minutes, during which time 6 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. (Use.)

5831J

Explanation:

Data;

Mass of water = 20kg

Height = 35m

Time = 14minutes

Mass if leakage = 6kg

g = 9.8 m

The bucket moves upward 35/14m/min at time t,

The height of the bucket =

y = (35 / 14) t metres above the ground.

Water drips out at the rate of 6kg / 14mins

At first, there was 20kg of water in the bucket, at time t, the mass remaining = ?

M = (20 - 6/14) t kg

Consider the time interval between t and t + ∇t. During this time, the bucket moves a distance of

∇y = (35/14 ∇t) J

W = lim ₓ→ₐ ∑ (20 - 3/7) t * g * 35/14 ∇t

W = ∫¹⁴₀ (20-3/7) t * g * 35/14 dt

W = 35/14g (20t - 3/14t²) |¹⁴

W = 35/14g [20 (14) - 3/14 (14) ²]

W = 35/14*g * (280-42)

W = 5831J