Question

Block A of mass 2.0kg and block B of mass 8.0kg are connected as shown by a spring of spring constant 80N/m and negligible mass. The system is being pulled to the right across a horizontal friction less surface by a horizontal force of 4.0 N, as shown, with both blocks experiencing equal constant acceleration.

(a) Calculate the force that the spring exerts on the 2.0 kg block.

(b) Calculate the extension of the spring. The system is now pulled to the left, as shown below, with both blocks again experiencing equal constant acceleration.

(c) Is the magnitude of the acceleration greater than, less than, or the same as before?

Greater Less ____ The same Justify your answer.

(d) Is the amount the spring has stretched greater than, less than, or the same as before?

Greater Less ____ The same Justify your answer.

(e) In a new situation, the blocks and spring are moving together at a constant speed of 0.50 m * to the left.

Block A then hits and sticks to a wall. Calculate the maximum compression of the spring.

Answer 1

(a) In a connected system, we must first find the acceleration of the system as a whole. The spring is internal when looking at the whole system and can be ignored.

Fnet = ma

(4) = (10)a

a = 0.4m/s2 → the acceleration of the whole system and also of each individual block when looked at separate

Now we look at just the 2kg block, which has only the spring force acting on its FBD horizontal direction.

Fnet = ma

Fsp = (2)(.4)

Fsp = 0.8N

(b) Use Fsp = k∆x 0.8 = (80) ∆x ∆x = 0.01 m

(c) Since the same force is acting on the same total mass and Fnet = ma, the acceleration is the same

(d) The spring stretch will be MORE. This can be shown mathematically by looking at either block. Since the 8 kg block has only the spring force on its FBD we will look at that one.

Fsp = ma k∆x = ma (80)(∆x) = (8)(0.4) ∆x = 0.04m

(e) When the block A hits the wall it instantly stops, then block B will begin to compress the spring and transfer its kinetic energy into spring potential energy. Looking at block B energy conservation:

Kb = Usp

mvb2 = k∆x2

(8)(0.5)2 = (80)∆x2

∆x = 0.16m