Question

A bag of cement originally weighing 134 lb was lifted at a constant rate. As it rose, cement also leaked out at a constant rate. By the time the bag had been lifted to 15 ft, 12 of the cement had leaked out. How much work was done lifting the cement this far? (Neglect the weight of the bag and lifting equipment.)

Answer 1

Answer:1920 lb-ft

Explanation:

Given

Weight of cement bag=134 lb

Cement leaked=12 lb

height =15 ft

Let Weight be a linear function of distance moved such that

w=-ax+b

at x=0 ,w=134 lb

at x=15, w=134-12=122 lb

Thus b=134 ,a=0.8

and work done is

$W=\int_{0}^{15}wdx$

$W=\int_{0}^{15}\left ( -0.8x+134\right )dx$

$W=\left ( -0.4\times x^2+134x\right )_0^15$

$W=-0.4\times 225+134\times 15$

W=1920 lb-ft

Answer 2

Answer:

Work done is 1.920 kJ

Solution:

As per the question:

The original weight of the sand = 134 lb

The final weight of the sand after the leakage of 12 lb of the sand from it = 134 - 12 = 122 lb

The height to which it was raised, h = 15 ft

Now,

The work done can be calculated as:

Average weight of the sand, $w_{a} = \frac{(134 + 122)}{2} = 128 lb$

Work done = $w_{a}\times h = 128\times 15 = 1920 J = 1.920 kJ$