Question

two identical blocks are connected by a spring (of negligible mass) with a stiffness of $39.7\frac{n}{m}$, and hung from a string that passes over a pulley to a larger block with mass $16kg$, as shown in the diagram. hanging like this stretches the spring from its equilibrium length by an amount of $0.21m$. the larger block rests on a scale. find the reading on the scale in newtons. provide at least one decimal place

Explanation:

Step1: Calculate spring force

Use Hooke's Law: $F_s = kx$
$F_s = 39.7\ \frac{N}{m} \times 0.21\ m$
$F_s = 8.337\ N$

Step2: Find mass of one small block

The spring force equals the weight of one small block: $F_s = m_s g$
$m_s = \frac{F_s}{g} = \frac{8.337\ N}{9.81\ \frac{m}{s^2}} \approx 0.8509\ kg$

Step3: Calculate total weight of small blocks

Total mass of two small blocks: $m_{total} = 2m_s$
$m_{total} = 2 \times 0.8509\ kg = 1.7018\ kg$
Total weight of small blocks: $W_s = m_{total}g$
$W_s = 1.7018\ kg \times 9.81\ \frac{m}{s^2} \approx 16.695\ N$

Step4: Calculate weight of large block

$W_l = m_l g$
$W_l = 16\ kg \times 9.81\ \frac{m}{s^2} = 156.96\ N$

Step5: Find scale reading

Scale reading = $W_l - W_s$
$Scale\ reading = 156.96\ N - 16.695\ N$

Answer:

140.3 N