That's ok I appreciate any help I can get. There were just two problems in this section that are giving me heart-ache and this happens to be one of them.
Hey, when I solve it using your expression I come up with -0.400 cm. Is that what you get as well?
If you take dV=0 then the final expression I come up with is 2kx=-.064, and when I solve for x, I get -.004m
A 2.00-kg ball is attached to the bottom end of a length of fishline with a breaking strength of (44.5 N) The top end of line is held stationary. The ball is released from rest with the line taut and horizontal (theta=90.0). At what angle theta(measured from the vertical) will the fishline...
In part (a) the speed of the projectile as it leaves the barrel is 1.40 m/s, and I found this out from taking the final energy=1/2 mv^2(final) subtracted from the initial energy=1/2kx^2(initial) and equating the change of energy to the negative of the frictional force*distance the projectile...
I don't need the height of the projectile after it leaves the cannon, just at what point inside the barrel of the cannon is the velocity the maximum. I know it's some point after the initial release from the spring because the longer the ball travels through the barrel the slower it is because...
Need Help Not the Answer Please!
A toy cannon uses a spring to project a 5.30-g rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel, and there is a constant force of...