Answer:
5831J
Explanation:
Data;
Mass of water = 20kg
Height = 35m
Time = 14minutes
Mass if leakage = 6kg
g = 9.8 m
The bucket moves upward 35/14m/min at time t,
The height of the bucket =
y = (35 / 14)t metres above the ground.
Water drips out at the rate of 6kg/ 14mins
At first, there was 20kg of water in the bucket, at time t, the mass remaining = ?
M = (20 - 6/14)t kg
Consider the time interval between t and t + ∇t. During this time, the bucket moves a distance of
∇y = (35/14 ∇t) J
W = lim ₓ→ₐ ∑ (20 -3/7)t *g* 35/14 ∇t
W = ∫¹⁴₀ (20-3/7)t * g * 35/14 dt
W = 35/14g (20t - 3/14t²) |¹⁴
W = 35/14g [20(14) - 3/14(14)²]
W = 35/14*g* (280-42)
W = 5831J
