Spring Forces Practice Test

A $0.250\ \text{kg}$ block is attached to two ideal springs on a horizontal frictionless table. Spring 1 has constant $k_1 = 120\ \text{N/m}$ and Spring 2 has constant $k_2 = 80.0\ \text{N/m}$. The springs are connected in parallel between a wall and the block so that both springs stretch or compress by the same amount x when the block is displaced. The block is pulled a small distance and released, oscillating about equilibrium.

Given values:

• $k_1 = 120\ \text{N/m}$
• $k_2 = 80.0\ \text{N/m}$
• $m = 0.250\ \text{kg}$

Forces and model: For a displacement x, each spring exerts a restoring force $F_{s1} = -k_1 x$ and $F_{s2} = -k_2 x$. The net restoring force is $$F_{\text{net}} = -(k_1 + k_2)x,$$ so Newton’s Second Law becomes $$m,\frac{d^2x}{dt^2} = -(k_1+k_2)x.$$

Refer to the system described above. What is the effective spring constant for the system?

A $2.0\ \text{kg}$ cart is attached to two springs connected in series along a horizontal frictionless track. Spring A has constant $k_A = 300\ \text{N/m}$ and spring B has constant $k_B = 600\ \text{N/m}$. The left end of spring A is fixed to a wall, spring A connects to spring B, and spring B connects to the cart. When the cart is displaced to the right from equilibrium by a small distance, both springs stretch, and each spring exerts a restoring force. For springs in series, the same force magnitude acts through both springs, and the total extension is the sum of individual extensions. The effective spring constant satisfies $\tfrac{1}{k_{\text{eff}}} = \tfrac{1}{k_A} + \tfrac{1}{k_B}$. The cart oscillates with small amplitude so that Hooke’s law applies.

Given values:

• Mass: m = $2.0\ \text{kg}$
• Spring constants: $k_A = 300\ \text{N/m}$, $k_B = 600\ \text{N/m}$
• Track is horizontal and frictionless

Refer to the system described above. What is the effective spring constant for the system?