At a construction site, a 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building. There is no appreciable friction on the box, since it is on roller bearings. The box starts from rest.
A. Find the acceleration of the bucket.
B. How fast is the bucket moving after it has fallen 1.50 m (assuming that the box has not yet reached the edge of the roof)?

If an object's velocity changes, it is said to have been accelerated. An object's velocity can alter depending on whether it moves faster or slower or in a different direction. A falling apple, the moon orbiting the earth, and a car stopped at a stop sign are a few instances of acceleration. Through these illustrations, we can see that acceleration happens whenever a moving object changes its direction or speed, or both.

What is Velocity?

The direction in which a body or an object is moving.

What are the calculations?

Given that the mass of bucket m = 22.0 kg

Weight of the box W = 375 N

Then the mass of box M = W / g

= 375 / 9.8 m/[tex]s^{2}[/tex]

= 38.27 N

Let T be the tension in the string

Let a be the acceleration of the bucket

Apply Newton's second law to the bucket

mg - T = ma

T = m (g - a) ....... (1)

Apply Newton's second law to the box

T = Ma ....... (2)

From equations (1) and (2) we get

Ma = m (g-a)

( m+ M )a = m g

a = m g / (M + m)

= (22.0 kg) (9.8 m/[tex]s^{2}[/tex]) / (22.0 kg + 38.27 kg )

= 3.58 m/[tex]s^{2}[/tex]

Initial velocity of the bucket u = 0

Let v be the final velocity of the bucket after displacement s = 1.5 m

From equation of motion

v2 - u2 = 2as

v2 - 0 = 2as

v = √2as

v = √2(3.58 m/[tex]s^{2}[/tex])(1.5 m)

v = 3.28 m/s

Hence, the acceleration of the bucket is 3.58 m/[tex]s^{2}[/tex] and the bucket is moving with a velocity of 3.28 m/s.

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