Explain why it is easier to lower the ramp by pulling on the rope rather than pulling on the handle - AQA - GCSE Physics - Question 10 - 2019 - Paper 1

To find the elastic potential energy (E), we first determine the force (F) acting on the spring using the moment equation:

$924 = F \times 0.15$

Calculating the force: $F = \frac{924}{0.15} = 6160 \text{ N}$

Next, we calculate the spring constant (k) using Hooke's law. Since the force is known and the spring is stretched by 0.250 m, we use:

$F = k \times x \Rightarrow k = \frac{F}{x} = \frac{6160}{0.250} = 24640 \text{ N/m}$

Now, we apply the formula for elastic potential energy:

$E = \frac{1}{2} k x^2 = \frac{1}{2} \times 24640 \times (0.250)^2$

Calculating the elastic potential energy:

$E = \frac{1}{2} \times 24640 \times 0.0625 = 770 \text{ J}$