Question

draw datum and free body diagram please !

The steel ingot has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m/s when it collides with the "rested" spring assembly. Determine the maximum deflection in each spring needed to stop the motion of the ingot. Take ka = 5 kN/m, kb = 3 kN/m. 0.5 m -0.45 m kg kA с B

Answer 1

Let us assume Spring 2 + 0.05 ingot comes to rest when Bis compressed by an_amount x' Hence, the..compression in__spring_A_will be at [0.5-0.45]" final energy of system when ingot comes to resto - KANA spring energy 1 * 1 x 5 (n+0.05)?+ 1*3 (72). + 2 2 Ef 402 +0.25 +0.0125 KN-m OR f KI 2 Initial e energy of system 2 2 Ei I my 2 x 1800 x 0.5² IN 2 0.225 kW-om, OR 225 CSScanned with CamScanner KJ

As all the forces present in system are conservative cam_apply "Conservation of energy"_principle We final Energy Initial energy 402 +0.25 +0.0125 = 0.225 2 Game + 0.25% + (-0.2125) 0 upon solving, X = -0.25 + 25)- 40 (-0.2125 2x 4 *, = 0.2 m. Therefore, compression in compression in 0.2 m Spring Ba A: spring 0.2+0.00 = 0.25 m To be more accurate compression in spring Ba spring 0. 20135 m A = 0o 25155 m. Esscanned with Camscanner compression in
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