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

W = 2352 J

Step-by-step 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* 3] dt`

`W= 3* 9.8* [10t-(0.4t^(2))/(2)]^(10)_(0)`

`W= 2352 J`