Question

A block of mass $5\text{ kg}$ moves along the floor of a hall with an initial speed of $5\text{ m s}^{-1}$. It strikes an uncompressed spring and compresses it till the block becomes motionless. If the force constant of the spring is $10000\text{ N m}^{-1}$ and the spring is compressed by $5\text{ cm}$, calculate the effective force of kinetic friction.

• A. $125\text{ N}$
• B. $16.4\text{ N}$
• C. $0\text{ N}$
• D. $18.7\text{ N}$

Hint:

The kinetic energy lost ($18.75\text{ J}$) goes into two accounts: spring storage ($12.5\text{ J}$) and friction dissipation. The friction work must equal the remainder ($18.75 - 12.5 = 6.25\text{ J}$). Dividing $6.25\text{ J}$ by the short displacement $0.05\text{ m}$ instantly yields $125\text{ N}$.

Answer 1

The Correct Option is A

Understanding the Concept: According to the Work-Energy Theorem, the net work done by all forces acting on a system equals its net change in kinetic energy: \[ W_{\text{spring}} + W_{\text{friction}} = \Delta K = K_{\text{final}} - K_{\text{initial}} \] The work done by a spring during compression is $-\frac{1}{2}kx^2$, and the work done by a uniform kinetic friction force over displacement $x$ is $-f_k \cdot x$.
Step 1: Set up the energy balance equation elements.
We are given:
Mass, $m = 1.5\text{ kg}$
Initial speed, $v = 5\text{ m s}^{-1}$
Spring constant, $k = 10000\text{ N m}^{-1}$
Compression distance, $x = 5\text{ cm} = 0.05\text{ m}$
Final speed, $v_f = 0\text{ m s}^{-1}$ (comes to rest)
Let's evaluate the initial kinetic energy: \[ K_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.5 \times 5^2 = 0.75 \times 25 = 18.75\text{ J} \]
Step 2: Calculate potential energy stored in the spring.
\[ U_{\text{spring}} = \frac{1}{2} k x^2 = \frac{1}{2} \times 10000 \times (0.05)^2 = 5000 \times 0.0025 = 12.5\text{ J} \]
Step 3: Apply the work-energy balance to solve for friction force ($f_k$).
\[ -\frac{1}{2}kx^2 - f_k \cdot x = 0 - K_{\text{initial}} \] \[ -12.5 - f_k \cdot (0.05) = -18.75 \] Rearranging terms: \[ f_k \cdot (0.05) = 18.75 - 12.5 = 6.25 \] \[ f_k = \frac{6.25}{0.05} = 125\text{ N} \]