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project 7 (lifting a leaky bucket).a bucket weighs 5 kg empty. it is lifted from the ground to a height of 10 m at aconstant speed with a rope weighing 1 kg/m (10 m of rope is lifted). the bucket starts with 20 kg of water, but water leaks out the bottom at a constant rate, withhalf gone by the end.how much work is done in lifting the bucket, water, and rope (in newton-meters)?

1 Answer

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Final answer:

The work done in lifting the bucket, water, and rope is 2450 J.

Step-by-step explanation:

The work done in lifting the bucket, water, and rope can be calculated by adding the work done for each component separately. To do this, we need to calculate the work done against the force of gravity for each component and the work done against the weight of the rope.

First, let's calculate the work done to lift the bucket. The bucket weighs 5 kg, so the work done to lift it 10 m is given by:

Work = force x distance = weight x height = (mass x gravity) x height = (5 kg x 9.8 m/s²) x 10 m = 490 J

Next, let's calculate the work done to lift the water. Half of the water leaks out, so the remaining water weighs 20 kg / 2 = 10 kg. The work done to lift this water 10 m is the same as the work done to lift the bucket, since both have the same weight:

Work = (10 kg x 9.8 m/s²) x 10 m = 980 J

Finally, let's calculate the work done to lift the rope. The rope weighs 1 kg/m and 10 m of rope is lifted, so the total weight of the rope is 1 kg/m x 10 m = 10 kg. The work done to lift this rope 10 m is:

Work = (10 kg x 9.8 m/s²) x 10 m = 980 J

The total work done is the sum of the work done for each component:

Total Work = 490 J + 980 J + 980 J = 2450 J

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