Question

An ideal spring of unstretched length 0.20 m is placed horizontally on a friction less table as shown. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0 kg. The 8.0kg block is also attached to a mass less string that passes over a small friction less pulley. A block of mass m = 4.0kg hangs from the other end of the string. When this spring-and-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0kg block is 0.70m above the floor.

 

(a) On the figures below, draw free-body diagrams showing and labeling the forces on each block when the system is in equilibrium. 

M = 8.0 kg m = 4.0 kg

(b) Calculate the tension in the string. 

(c) Calculate the force constant of the spring. 

The string is now cut at point P. 

(d) Calculate the time taken by the 4.0kg block to hit the floor. 

(e) Calculate the maximum speed attained by the 8.0kg block as it oscillates back and forth

Answer 1

(b) Simply isolating the 4kg mass at rest. 

F_net = 0 

F_t – mg = 0 

F_t = 39N

(c) Tension in string is uniform throughout, now looking at the 8 kg mass, F_sp = F_t = k∆x 

39 = k (0.05) 

k = 780 N/m

(d) 4 kg mass is in free fall. 

D = v_it + ½gt² – 0.7 = 0 + ½(– 9.8)t² 

t = 0.38 sec

(e) The 8 kg block will be pulled towards the wall and will reach a maximum speed when it passes the relaxed length of the spring. At this point all of the initial stored potential energy is converted to kinetic energy 

U_sp = K 

½ k ∆x² = ½ mv² 

½ (780) (0.05) = ½ (8)v²

v = 0.49m/s