Water weighing 42 kg is raised uniformly in a cylindrical bucket with a diameter of 15 cm.

Water weighing 42 kg is raised uniformly in a cylindrical bucket with a diameter of 15 cm. During 15 s of ascent, the speed is changed from 12 m / s to 25 m / s. Find the pressure on the bottom of the bucket.

Given:

m = 42 kilograms is the mass of water;

d = 15 centimeters = 0.15 meters – bucket diameter;

v1 = 12 m / s – the initial speed of the bucket;

v2 = 25 m / s – the final speed of the bucket;

t = 15 seconds – the time during which the bucket speed changed;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine P (Pascal) – the water pressure at the bottom of the bucket.

Let’s find the acceleration with which the water is raised:

a = (v2 – v1) / t = (25 – 12) / 15 = 13/15 = 0.9 m / s ^ 2.

Then, since the bucket is being lifted up, the weight of the water will be:

F = m * (g + a) = 42 * (10 + 0.9) = 42 * 10.9 = 457.8 Newtons.

The area of ​​the bottom of the bucket will be equal to:

S = pi * d ^ 2/4 = 3.14 * 0.15 ^ 2/4 = 0.02 m ^ 2.

Then the water pressure at the bottom of the bucket will be equal to:

P = F / S = 457.8 / 0.02 = 22890 Pascal = 22.3 kPa.

Answer: the water pressure at the bottom of the bucket is 22.3 kPa.