Answer:
work done is 470400 J
Explanation:
given data
long = 3 m
wide = 2 m
deep = 4 m
density = 1000 kg/m³
to find out
Find the amount of work in joules
solution
we know here volume = area × length
so
area = 0.5 ×base×height
area = 0.5 ×2×4
area = 4 m²
so volume = 4 × 3 = 12 m³
and
weight of water = volume ×density×g
weight = 12 × 1000 × 9.8
weight = 117600 N
so
work done = weight × height
work done = 117600× 4
work done = 470400 J
To solve this problem it is necessary to apply the concepts related to energy conservation. Therefore, the work done will initially be equivalent to the change in kinematic energy. And this kinematic energy will be equivalent to the sum of energy (work) carried out by force and friction. In this way we have to
Now,
For each one we have
Then at the equation of equilibrium of energy we have,
Therefore the work done by friction on the chair is -833.7J
<span>C. The mass of box A is one–third the mass of box B.</span>