Question

A bucket of water of mass 30 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 12 minutes, during which time 4 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 g=9.8m/s^2)

Answer 1

Step-by-step explanation:

Given that:

mass of the bucket, M = 10 kg

velocity of pulling the bucket, v = 3m.min^{-1}m.min

−1

height of the platform, h = 30 m

time taken, t = 10 min

rate of loss of water-mass, m = 0.4 kg.min^{-1}0.4kg.min

−1

Here, according to the given situation the bucket moves at the rate,

v=3 m.min^{-1}v=3m.min

−1

The mass varies with the time as,

M=(10-0.4t) kgM=(10−0.4t)kg

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x = 3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

For work calculation:

W=\int_{0}^{10} [(10-0.4t).g\times 3] dtW=∫

0

10

[(10−0.4t).g×3]dt

W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}W=3×9.8×[10t−

2

0.4t

2

]

0

10

W= 2352 JW=2352J