Two blocks are at rest on a horizontal frictionless surface with a compressed spring between them.The 10-kg block accelerates when the string holding the system in compression is cut and the 20-kg box accelerates in the opposite direction.What is the magnitude of the force from the expanding spring on the 20-kg box when the 10-kg box is undergoing an acceleration of 8.0 ?
A)40 N
B)160 N
C)(160/3)N
D)80 N

Question

Two blocks are at rest on a horizontal frictionless surface with a compressed spring between them.The 10-kg block accelerates when the string holding the system in compression is cut and the 20-kg box accelerates in the opposite direction.What is the magnitude of the force from the expanding spring on the 20-kg box when the 10-kg box is undergoing an acceleration of 8.0  ?
A)40 N 
B)160 N 
C)(160/3)N 
D)80 N

Quizplus · Accepted Answer

We can use the conservation of momentum to find the magnitude of the velocity of each block after the spring is released:

m1v1 = m2v2

10kg (0 m/s) = 20 kg (v2)

v2 = 0 m/s

This means that the 20-kg block does not move initially and only starts moving because of the force from the expanding spring.

Next, we can use Newton's second law to find the magnitude of the force on the 10-kg block:

F = ma

F = 10 kg x 8.0 m/s^2

F = 80 N

Since the spring is compressed between the blocks, the force on the 10-kg block is equal and opposite to the force on the 20-kg block. Therefore, the magnitude of the force from the expanding spring on the 20-kg box is also 80 N, which corresponds to choice D.

Two Blocks Are at Rest on a Horizontal Frictionless Surface