Question

The mass of a bucket full of water is 15 kg. It is
being pulled up from a 15m deep well. Due to a
hole in the bucket 6 kg water flowsout of the bucket
The work done in drawing the bucket out of the well
will be (g =10m/s2)​

Answer 1

Assume the water uniformly loses from bucket, i.e., equal amount of water loses for a same interval of time.

• 6 kg water loses when moving 15 m upwards.

• $\displaystyle\sf{\dfrac{6}{15}=\dfrac{2}{5}\ kg}$ water loses when moving 1 metre upwards.

• $\displaystyle\sf{\left(\dfrac{2}{5}\ x\right)\ kg}$ water loses when moving x metre upwards.

The force required to pull the bucket when it is at x distance above from the bottom of the well is equal to its weight at that moment, i.e.,

$\displaystyle\longrightarrow\sf{F=\left(15-\dfrac{2}{5}\ x\right)g}$

Work done to pull it through a small distance 'dx' is,

$\displaystyle\longrightarrow\sf{dW=\left(15-\dfrac{2}{5}\ x\right)g\ dx}$

Total work done to pull it from the bottom (x = 0 m) to the top (x = 15 m) is,

$\displaystyle\longrightarrow\sf{W=g\int\limits_0^{15}\left(15-\dfrac{2}{5}\ x\right)\ dx}$

$\displaystyle\longrightarrow\sf{W=g\left[\int\limits_0^{15}15\ dx-\dfrac{2}{5}\int\limits_0^{15}x\ dx\right]}$

$\displaystyle\longrightarrow\sf{W=g\left[15\big[x\big]_0^{15}-\dfrac{2}{5}\left[\dfrac{x^2}{2}\right]_0^{15}\right]}$

$\displaystyle\longrightarrow\sf{W=g\left[15(15-0)-\dfrac{1}{5}\left(15^2-0^2\right)\right]}$

$\displaystyle\longrightarrow\sf{W=g\left[225-45]}$

$\displaystyle\longrightarrow\sf{\underline{\underline{W=180g\ J}}}$

Taking $\displaystyle\sf{g=9.8\ m\ s^{-2},}$

$\displaystyle\longrightarrow\sf{\underline{\underline{W=1764\ J}}}$

If $\displaystyle\sf{g=10\ m\ s^{-1},}$

$\displaystyle\longrightarrow\sf{\underline{\underline{W=1800\ J}}}$