Physics Question: Nonconservative forces and conservation of energy?

Question

At a carnival, you can try to ring a bell by striking a target with a 9.00-kg hammer. In response, a 0.400-kg metal piece is sent upward toward the bell, which is 5.00 m above. Suppose that 25.0% of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

Please show your work with all formulas used. (conservation of energy formula I believe)
Show your answer in terms of the variables given and a numerical answer.

Thanks.

Answer 1

Figure out the energy diffece needed to be imparted to the piece to just strike the bell first. Gravitational potential energy is : U=mgh

so,
0.4kg * 5m *g = 19.6J, which means we need the hammer's kinetic energy to be 19.6J / 0.25 (since only 25% of the hammer's kinetic energy is used to send the bell upward) = 78.4N

Kinetic energy is 1/2mv^2 = 78.4N, and we know the m of the hammer is 9kg, so we rearrange the formula to get v^2 = 2*78.4J/9kg, so v = sqrt(2*78.4/9) = 4.17m/s

Check it: does 0.5*9kg*(4.17m/s)^2 = 0.25*0.4kg*5m*9.8m/s^2

It does, so the our method works, 1/2mv^2 (for the hammer) = mgh/0.25 (for the piece)

Answer 2

KEL = launch KE = Fd = mad; where F is the force sending the metal piece d meters upward. KEL = 1/4 KEI = 1/4 impact KE = 1/4 (1/2) M v^2; where M = 9 kg, d = 5 meters, m = .4 kg, and a = g = 9.81 m/sec^2, which is the rate of deceleration of the metal piece due to gravity

Thus, mgd = 1/4 (1/2) M v^2; so that v = sqrt(8mgd/M). You can do the math.

Lesson learned: The energy of the rising metal piece is exactly the work to raise its weight (mg) the d = 5 meters. Since it was given that that work mgd is 1/4 of the work (energy) put into the hammer, we equated mgd = 1/4 KEI = 1/4 (1/2) Mv^2, which is where the 8 came from in v = sqrt(8mgd/M).

By the way, I assumed a frictionless track for the metal piece. Had there been friction, that would have required more v and energy to reach the d height.

Answer 3

The vector is the final one million/2 of a parabola the place the x value is the linear velocity of the rock at 25 m/s, and maintains to be consistent because of the fact we don't have a coefficient of friction for air resistance. The y value is damaging, because of the fact the rock starts falling because of gravity at 32ft/s/s as quickly as that's launched. The rock would be vacationing down on the comparable velocity as though it have been dropped vertically.