A2.0-kg mass and a 3.0-kg mass are on a horizontal frictionless surface, connected by a massless spring with spring constant k = 140 n/m. a 15-n force is applied to the larger mass, as shown in fig. 4.23. how much does the spring stretch from its equilibrium length?

The solution would be like this for this specific problem:

=> F/2 = (15 - F)/3 
=> 3F = 30 - 2F 
=> 5F = 30 
=> F = 6 N 

So if the spring stretches by x, 
F = kx 
=> x = F/k = 6 / 140 m = 600 / 140 cm = 4.3 cm

The spring stretches for 4.3 cm.

I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

and 1and 1

the total distance traveled by the car is

30km + 49km = 79 km .