A bucket full of water weighs 5 kg , it is pulled with constant speed from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate of 0.2 kg m- 1 . Ignore the force due to leakage. The total work done in pulling the bucket up from the well is (g = 10 m s- 2)

Question

A bucket full of water weighs 5 kg , it is pulled with constant speed from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate of 0.2 kg m- 1 . Ignore the force due to leakage. The total work done in pulling the bucket up from the well is (g = 10 m s- 2) (A) 600J (B) 400J (C) 100J (D) 500J

Tardigrade Admin · Accepted Answer

The weight of bucket when it has been pulled up a distance

x

is

(5 - 0.2 x)

. Hence, the required work is

W = \int_{x = 20}^{x = 0} (5 - 0.2 x) \times 10 \times dx

= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}

W = 50 \times 20 - (20)^{2} = 600 J

$600 J$

$400 J$

$100 J$

$500 J$

The weight of bucket when it has been pulled up a distance $x$ is $(5 - 0.2 x)$ . Hence, the required work is
$W = \int_{x = 20}^{x = 0} (5 - 0.2 x) \times 10 \times dx$ $= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}$ $W = 50 \times 20 - (20)^{2} = 600 J$

Q. A bucket full of water weighs 5kg , it is pulled with constant speed from a well 20m deep. There is a small hole in the bucket through which water leaks at a constant rate of 0.2kgm−1 . Ignore the force due to leakage. The total work done in pulling the bucket up from the well is (g=10ms−2)

600J

400J

100J

500J

The weight of bucket when it has been pulled up a distance x is (5−0.2x). Hence, the required work is W=∫x=20x=0​(5−0.2x)×10×dx =[50x]x=0x=20​−[22x2​]x=0x=20​ W=50×20−(20)2=600 J

$600 J$

$400 J$

$100 J$

$500 J$

The weight of bucket when it has been pulled up a distance $x$ is $(5 - 0.2 x)$ . Hence, the required work is
$W = \int_{x = 20}^{x = 0} (5 - 0.2 x) \times 10 \times dx$ $= [50 x]_{x = 0}^{x = 20} - [2 \frac{x ^{2}}{2}]_{x = 0}^{x = 20}$ $W = 50 \times 20 - (20)^{2} = 600 J$