A 2.0-kg box and a 3.0-kg box on a perfectly smooth horizontal floor have a spring of force constant 250 N/m compressed between them. If the initial compression of the spring is 6.0 cm, find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

Solution 36E The force required to stretch a spring is F = kx, where k is the spring constant and x is the elongation. Let us now have a look at the free body diagram as shown below. Given, force constant k = 250 N/m 2 Compression x = 6.0 cm = 6.0 × 10 m 2 Therefore, the force required to stretch the spring is F = kx = 250 N/m × 6.0 × 10 m F = 15 N Now, mass of the 1st box m = 2.0 kg 1 So, its acceleration is a = F/m 1 1 a 1 15 N/2.0 kg 2 a 1 7.5 m/s When released, the acceleration of the 2.0 kg box will be 7.5 m/s . The mass of the 2nd box is m = 3.0 kg 2 So, its acceleration a = 15 N/3.0 kg 2 2 a 2 5.0 m/s When released, the acceleration of the 3.00 kg box will be 5.0 m/s . 2