Water falls without splashing at a rate of `0.250 Ls^(-1)` from a height of 2.60 m into a 0.750kg bucket on a scale. If the bucket is originally empty, what does the scale read 3s after water starts to accumulate in it?

Explanation

After 3s of pouring the bucket contains `(3s)(0.25 Ls^(-1))=0.75L` of water, with mass `0.75 L xx (1kg//1L)=0.75 kg`, and feeling gravitational force `0.75kg(9.8ms^(-2))=7.35N`. The scale through the bucket must exert `7.35N` upward on this stationary water to support its weight. The scale must exert another `7.35N` to support the `0.75 kg` bucket itself.
water is entering the bucket with the speed given by
`mgy_("top")=(1//2)mv_("impact")^(2)`
`v_("impact")=(2gy_("top"))^(1//2)=[2(9.8 ms^(-2))2.6 m]^(1//2)`
`=7.14 ms^(-1)` downward.
The scale exerts an extra upward force to stop the downward motion of this additional water, as described by
`mv_(i mpact)+f_(e xtra)=mv_(f)`
`(dm//dt)v_("impact")+F_(extra)-0`
`F_(extra)=-(dm//dt)v_(impact)=-(0.25 kg s^(-1))(-7.14 ms^(-1))`
`=+1.78 N`
altoghther the scale must exert,
`7.35N+7.35 N+1.78N=16.5N`.