So is this what you're saying I should do?
Find the PE=mgh=(20)(9.8)(20)=784 J
And find the KE=(1/2)(20kg)(4.0^2)=3200 J
And then add PE+KE so that my final answer is 784+3200=3894?
Because the final answer is supposed to be 4.08kN*m
Homework Statement
You carry a 7.0-kg bag of groceries 1.2 m above the level floor at a constant velocity of 75 cm/s across a room that is 2.3 m wide. How much work do you do on the bag in the process?
Homework Equations
KE=(1/2)(m)(v^2)
The Attempt at a Solution
I haven't a clue where to...
I solved for F in my equation of W=Fd by using the kinetic energy formula. I did (1/2)*(20)*(4^2) and got 160J. I plugged that into the W=Fd as my F. So then I solved doing F=160J and d=20m but I got 3200 N*m instead and that's not the right answer. The answer is 4.08kN*m apparently
Homework Statement
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 20.0-m-tall building. The bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the building. What is the minimum amount of work that the worker did in lifting the...