A steel part with a volume of 0.4 m3 and a density of 7800 kg / m3 is lifted evenly and to a height of 15 meters

A steel part with a volume of 0.4 m3 and a density of 7800 kg / m3 is lifted evenly and to a height of 15 meters in 30 seconds. Find: a) body weight; b) the force of gravity acting on the body; c) body weight; d) work on lifting the body; e) the power developed at the same time; f) lifting speed; g) body impulse; h) potential energy at the top point; i) kinetic energy during ascent; j) * buoyancy when immersed (kerosene density 800 kg / m3); l) * body weight in kerosene.

Given:
ρ1 = 7800kg / m ^ 3,
ρ2 = 800kg / m ^ 3,
V = 0.4m ^ 3,
h = 15m,
t = 30s;
a) Body weight:
m = ρ1 * V = 7800kg / m ^ 3 * 0.4m ^ 3 = 3120kg;
b) Gravity:
P = m * g,
where
g = 10m / s ^ 2 – acceleration of gravity,
F = 3120kg * 10m / s ^ 2 = 31200H = 31.2kN;
c) Body weight:
Since the body is lifted evenly, without acceleration, it means that its weight is equal to the force of gravity:
P = F = 31.2kN;
d) Work on lifting the body:
A = F * h = 31200N * 15m = 468000J = 468kJ;
e) Developed capacity:
N = A / t = 468000J / 30s = 15600W = 15.6kW;
f) Lifting speed:
v = h / t = 15m / 30s = 0.5m / s;
g) Body impulse:
p = m * v = 3120kg * 0.5m / s = 1560kg * m / s;
h) Potential energy at the top point:
Wp = m * g * h = 3120kg * 10m / s ^ 2 * 15m = 468000J = 468kJ;
i) Kinetic energy during ascent:
Wk = m * (v ^ 2) / 2 = 1560kg * 0.25 (m / s) ^ 2 = 390J;
j) Buoyancy in kerosene:
The buoyancy force is the force of Archimedes:
F_A = ρ2 * g * V = 800kg / m ^ 3 * 10m / s ^ 2 * 0.4m ^ 3 = 3200Н;
l) Body weight in kerosene:
It is determined by the difference between the gravity and Archimedes forces acting on the body:
P1 = F – F_A = 31200N – 3200N = 28000N = 28kN.