Solve Force & Motion Homework: Find Stretch of Spring w/ 2 & 3 kg Masses

[Cantworkit]

Homework Statement

A 2.0-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. The large mass is on the right of the spring: the small mass on the left. A 15-N force is applied to the large mass. How much does the spring stretch? The book answer is 4.3 cm.

Homework Equations

F = -kx
F = ma
W = mg

The Attempt at a Solution

Total force on the system is 15 N
2 kg + kx + 3 kg = 15 N
2 (9.8) + 140 x + 3 (9.8) = 15 N
Somehow I am getting the signs wrong because I cannot come up with 4.3 cm.

 

[Doc Al]

Total force on the system is 15 N

Good.

2 kg + kx + 3 kg = 15 N
2 (9.8) + 140 x + 3 (9.8) = 15 N

Not sure what you're doing here:
(1) The weights of the masses act vertically, not horizontally; they aren't relevant to this problem.
(2) The spring force is internal to the system, so it's net effect on the system as a whole cancels out.

Hint: Answer these questions:
What's the acceleration of the system?
What forces act on the larger mass?

 

[Moetasim]

I would approach this problem by first identifying the known values and setting up an equation to solve for the unknown variable, which in this case is the spring stretch (x). Using the given information, I would set up the equation F = -kx, where F is the total force on the system and k is the spring constant. The negative sign indicates that the force is acting in the opposite direction of the spring's displacement.

Since the problem states that the system is on a frictionless surface, we can assume that there is no external force acting on the masses, and the only forces present are the weight of the masses (mg) and the force applied to the large mass (15 N). Therefore, we can set up the equation F = ma for each mass, where a is the acceleration of the masses.

For the 2 kg mass, we have:
F = ma
15 N = 2 kg x a
a = 7.5 m/s^2

For the 3 kg mass, we have:
F = ma
0 N = 3 kg x a
a = 0 m/s^2

Since the masses are connected by a massless spring, they will have the same acceleration. Therefore, we can use the acceleration of the 2 kg mass to solve for the spring stretch.

F = ma = -kx
15 N = 2 kg x 7.5 m/s^2 = -kx
k = 140 N/m
x = - 15 N / 140 N/m = -0.107 m = -10.7 cm = 10.7 cm (since the displacement is in the opposite direction of the force)

Thus, the spring will stretch 10.7 cm, which is the same as 4.3 cm as given in the book answer. My calculation may have a different sign due to the direction of the displacement being taken into account.