Work and Kinetic Energy Problem for Pump

[marckc22]

Homework Statement

A pump is required to lift 800 kg. ( about 200 gallons) of water per minute from a well that is 14.0 m deep and eject it with a speed of 30 m/s
a.) how much work is done per minute in lifting the water?
b.) how much in giving it kinetic energy?

Homework Equations

W = Fd
F = ma
a = V/t
KE = 0.5(m)v^2

The Attempt at a Solution

a = v/t = (30 m/s)/(60s) = 0.5 m/s^2

F = ma = 800(0.5) = 400 N

W = (400 N)(14 m) = 5600 J

My answer for letter a. W = 5600 J

KE = 0.5(m)v^2 = 0.5(800)(30) = 12000 J

My answer for letter b. KE = 12000 J

Am i right? please let me know if there are any mistakes. Thank you...

 

[amg!]

a) $$W = mgh+ \frac{1}{2} mv^2 = 800kg*14m*9.81ms^{-2}+\frac{1}{2}*800kg*(30m/s)^2 = 469872J$$
b) $$\frac{1}{2}mv^2 = \frac{1}{2}*800kg*(30m/s)^2 = 360000J$$

 

[hage567]

amg!,

Please read PF guidelines about providing help in the Homework Forums.

https://www.physicsforums.com/showthread.php?t=5374