Force On Object With Constant Speed

ptguard1
Messages
13
Reaction score
0
A sack of flour of mass 6.00 kg is lifted vertically at a constant speed of 4.00 m/s through a height of 18.0 m

How great a force is required?

F=ma


I found the force that the sack has on the Earth (58.8N), but don't understand how I am suppose to find what force is being applied to the sack in order to give it a constant speed of 4.00 m/s.
 
Physics news on Phys.org
I figured out that the force is equal to the 58.8 N of force that the sack had on the earth. My logic was that an object can be lifted at different speeds with the same force, is this true or only partially true?
 
As long as the speed of the object is constant the total forces on the object should add up to 0. Since the gravitational field is pretty much constant at the altitude differences you have in your problem you just need to add the same force as the gravitational force.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top