Question

A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 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 ????=9.8m/s2.)

Answer 1

Answer:

W = 2352 J

Explanation:

Given that:

• mass of the bucket, M = 10 kg

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

• height of the platform, h = 30 m

• time taken, t = 10 min

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

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

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

The mass varies with the time as,

$M=(10-0.4t) kg$

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:

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

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

$W= 2352 J$