1. **State the problem:** We need to find the force applied at the end of the handle of a screw jack to lift a load of 2000 N. 2. **Given data:** - Pitch of screw jack, $p = 0.5$ cm - Length of handle, $r = 2$ cm - Load, $W = 2000$ N - Efficiency, $\eta = 35\% = 0.35$ 3. **Formula used:** The ideal mechanical advantage (IMA) of a screw jack is given by: $$\text{IMA} = \frac{2\pi r}{p}$$ The actual mechanical advantage (AMA) considering efficiency is: $$\text{AMA} = \eta \times \text{IMA}$$ Force applied $F$ is related to load $W$ and AMA by: $$F = \frac{W}{\text{AMA}}$$ 4. **Calculate IMA:** $$\text{IMA} = \frac{2 \pi \times 2}{0.5} = \frac{4\pi}{0.5} = 8\pi \approx 25.13$$ 5. **Calculate AMA:** $$\text{AMA} = 0.35 \times 25.13 = 8.7955$$ 6. **Calculate force applied $F$:** $$F = \frac{2000}{8.7955} \approx 227.4 \text{ N}$$ **Final answer:** The force that must be applied at the end of the handle is approximately $227.4$ N.