Help on energy conservation?

i need help answering this physics question plz its about work and conservation of energy

• a partially filled bag of cement is tossed off of bridge with an initial upward velocity of 3.0 m/s. the bag has a mass of 16 kg and falls 40.0 m to the river below. determine the velocity of the bag just before striking the water. i found that and it comes out to be 28.2 m/s ( correct me if i am wrong plz) b. imagine that water were replaced with a trampoline . the trampoline has a spring constant of 2000 N/m. determine the distance the bag will sink in the trampoline before coming to a stop.? i know that i have to use kinetic energy and potential energy and elastic potential energy but i dont know wt to set = to zero and what not.. can someone plz help me !

• Answer:

  Hi again, pleased to meet you. Solving this question can be done in a single step, or cutting it into simpler problems. The second way is better for learning it. Let's go: From the time t=0 when the bag is pushed up until it stops moving upwards. We need to know how high it will go. Lets denote this height as Y. And lets count that height from the bridge level. Vo=3 m/s V=0 a= g= 9.81 m/s² Kinetik energy: Ek= 1/2 m v² Potential Energy: Ep= mgY Both must balance, then ........ Ek=Ep 1/2 m V²= mgY; 1/2 V²= gY Y= 1/2 V²/g Y= 0.458 meters up. Now the bag starts falling. New height is total falling height, the one the bag went up, plus the bridge height. The height of the bridge over the river, will be denoted as Yo. We'll balance Energies again. Ep= m g (Y+Yo) Ek= 1/2 m V² Ep=Ek 1/2 m V² = m g (Y+Yo); 1/2 V² = g (Y+Yo) V²= 2 g(Y + Yo) V= √[2 g(Y + Yo)] V= 28.17 m/s YOU ARE RIGHT ¡¡ Lets now go with the trampoline. Lets denote the elastic constant as K. Lets denote the sinking distance of the trampoline as Yb Force needed for pushing the trampoline a distance Yb F= K*Yb Work needed for that skinking, starting from the leveled possition (trampoline at rest). W= F * Yb W= K * Yb * Yb W= K * Yb² As the trampoline starts from the resting possition, the Energy levels difference between the initial state and the bent one, will be equal to the work computed above. Energy at rest; Eo=0 Bent energy: E= K * Yb² Work= difference between both energy states. W= E-Eo= K * Yb² Hence ....... Potencial Energy absorbed by the trampoline during its compression, denoted by Eb is equal to the bending work required: Eb= K * Yb² That will balance the Kinetic energy of the falling bag. Ek=Eb 1/2 m V²= K * Yb² Speed of the bag: V= 28.17 m/s Mass of the bag: m= 16 Kg Trampoline constant: K= 2000 N/m Yb²= (m V²)/2K Yb= √[(m V²)/2K] Yb= 1.78 m. The bag will make the trampoline sink 1.78 meters.