the_d Messages 127 Reaction score 0 Thread starter Oct 28, 2006 #1 My problem is to find the force needed to give the system of bodies a velocity of 30 ft/s after moving 20 ft. from rest. Can anyone help me with this? All i have done is found EFx and EFy for the 50lb weight Attachments p1.JPG 11.1 KB · Views: 381 Last edited: Oct 28, 2006
My problem is to find the force needed to give the system of bodies a velocity of 30 ft/s after moving 20 ft. from rest. Can anyone help me with this? All i have done is found EFx and EFy for the 50lb weight
radou Homework Helper Messages 3,149 Reaction score 8 Oct 28, 2006 #2 Didn't you post this question earlier? (Or was it someone else with the same question?) As said, use the fact that the work done by that force on the distance of 20 ft equals the change of kinetic energy.
Didn't you post this question earlier? (Or was it someone else with the same question?) As said, use the fact that the work done by that force on the distance of 20 ft equals the change of kinetic energy.
radou Homework Helper Messages 3,149 Reaction score 8 Oct 28, 2006 #4 the_d said: is there an equation for that?? If you know the definition of work, and the definition of kinetic energy, you should be able to write down this equation yourself.
the_d said: is there an equation for that?? If you know the definition of work, and the definition of kinetic energy, you should be able to write down this equation yourself.
the_d Messages 127 Reaction score 0 Oct 28, 2006 #5 so W = KE, where does the coeffeficient of kinetic friction play in all this??
radou Homework Helper Messages 3,149 Reaction score 8 Oct 28, 2006 #6 the_d said: so W = KE, where does the coeffeficient of kinetic friction play in all this?? You didn't mention any kinetic friction coefficient.
the_d said: so W = KE, where does the coeffeficient of kinetic friction play in all this?? You didn't mention any kinetic friction coefficient.
radou Homework Helper Messages 3,149 Reaction score 8 Oct 28, 2006 #9 the_d said: does that play a role?? Yes, it does, because you have to include the force of friction into your sum of forces acting on the object.
the_d said: does that play a role?? Yes, it does, because you have to include the force of friction into your sum of forces acting on the object.