Question

3. [7] The two blocks shown below have masses m = 1.0 kg and m2 = 3.0 kg. Assume that 0 = 25°, k = 16 N/m, and H=0.11 between mi and the incline. Neglect friction in the pulley and assume that the pulley and string and cable are massless. If this system is released initially at rest with the spring in its equilibrium position (no stretch), what is the speed of m2 after it has descended 20 cm? EESTELE SELECT ONLARI LO AND THE FRUM RESEP HERE ERRELL Par PER FIL D ELSE THESEK m . m2 Елім- PASTELERID SELLEEN TELES LEECEEEEEEE

using the conservation of energy

Answer 1

K 0-25 - A'Y .2 m oozsinas 0 mgsius mi 1 K Mgos 25 Soln - Given m, = 1 kg m₂ = 3kg K= 16N/m, Mx = 0-11 h = 20cm = 0.2m L = 0.2m If My moves clown by 0.2m then mi also moves 0.2m on the inclined surface during this motion potential energy of the block me will increase and some part of energy will dissipate to overcome FBD friction force also some part of energy will stored in the spring as spring potential energy, as the spring will Streched by 0.2m from FBD of block m, Normal force N=m, glos 25° friction force f = M, N = Hung Cos 25° change in potential energy of block M₂ = magh change in potential energy of block m = migy change in potential energy in spring = 1 k 22 mig

Energy dissipated to overcome friction force = fxL change in kinetic energy of block m₂ = 1 m2 V². V₂ is the speed of me after it has decreased 20cm. Now by Conservation of energy we can balance energy as given below magh = migy + 1 KL? + fxL + 1 m2 V² 3x9.81% 0-2 = 149.81*(0-2xsin25") + 4x16x(0.232 + 0.ll x \x 9.81* Cos 25°x [V = 1.74 m/s 3 x y z