Question

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is originally empty, what does the scale read (in newtons) 3.20 s after water starts to accumulate in it

Answer 1

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

Answer 2

Answer:

F_scale ≈ 15.12 N

Explanation:

We are given;

Mass flow rate; m' = 0.2 l/s

Time; t = 3.2 s

Mass of bucket; m_b = 0.730 kg

Height; h = 3.6 m

Now, mass of water is;

m_w = 0.2 l/s × 3.2 s

m_w = 0.64 l

From conversion, 1 litre = 1 kg

Thus: m_w = 0.64 kg

Now, let's calculate final velocity from Newton's third equation of motion.

v² = u² + 2gh

Initial velocity is 0. Thus;

v² = 0 + 2(9.8 × 3.6)

v² = 70.56

v = √70.56

v = 8.4 m/s

Now, total mass of water and bucket is;

m_t = m_w + m_b = 0.64 + 0.73

m_t = 1.37 kg

Force on the scale is calculated from;

F_scale = (m_t)g + (m_w)v/t

F_scale = (1.37 × 9.81) + (0.64 × 8.4/3.2)

F_scale ≈ 15.12 N