Question

b) The cylinder in Fig. B1b has a mass of 20 kg and is released from rest when h 0. Determine its speed when h 3 m. Each spring has a stiffness k 40 N/m and an unstretched length of 2 m. 2 m Fig. B1b

Answer 1

Initial energy of the system (E_i) = 0
Final energy of the system (E_F)= Spring potential energy + Gravitational potential energy + Kinetic energy of cylinder
Now the extension in the spring
x= Final length - Initial length
x = (3² +2²)^1/2 - 2 = 1.61 m
Now putting the value
E_F = 2*[(1/2)kx²] + mg(-h) + (1/2)mv²
where m is mass of cylinder = 20 kg
h is height by which cylinder is lowered = 3 m
k is spring stiffness = 40 N/m
v is speed of cylinder
E_F = 2[(1/2)*40*1.61²] + [20*9.81*(-3)] +[(1/2)*20*v²]
E_F= 103.684 - 588.6 +10v²
Now taking energy conservation
Final energy = Initial energy
103.684 - 588.6 +10v² = 0
10v² = 484.916
v = 6.964 m/s
Hence the final speed of the cylinder is 6.964 m/s