Question

20 a spring stretches by 0.17 m when a bag of potatoes weighing 55 n is hung from its end. if the spring constant is 323.5 n/m how much elastic potential energy does the spring have when it is stretched this far?
○ 18.7 j
○ 9.3 j
○ 2.3 j
○ 4.7 j

Explanation:

Step1: Recall elastic PE formula

The formula for elastic potential energy is $U = \frac{1}{2}kx^2$, where $k$ is spring constant, $x$ is displacement.

Step2: Substitute given values

Given $k=323.5\ \text{N/m}$, $x=0.17\ \text{m}$

$$ U = \frac{1}{2} \times 323.5 \times (0.17)^2 $$

Step3: Calculate step-by-step

First compute $(0.17)^2 = 0.0289$
Then $\frac{1}{2} \times 323.5 = 161.75$
Finally $161.75 \times 0.0289 \approx 4.7\ \text{J}$

Answer:

D. 4.7 J