1. **Problem statement:** A load weighing 120 N is raised a distance of 4 cm by a machine. The worker exerts a force of 50 N through a distance of 12 cm. We need to find: a) Input work b) Ideal Mechanical Advantage (IMA) c) Efficiency of the machine 2. **Formulas given:** - Input work: $W_i = F_i D_i$ - Output work: $W_o = F_o D_o$ - Actual Mechanical Advantage (AMA): $\text{AMA} = \frac{F_o}{F_i}$ - Ideal Mechanical Advantage (IMA): $\text{IMA} = \frac{D_i}{D_o}$ - Efficiency: $\text{Eff} = \frac{W_o}{W_i} = \frac{\text{AMA}}{\text{IMA}}$ --- 3. **Step a) Calculate input work $W_i$:** Given: $F_i = 50$ N (input force) $D_i = 12$ cm = 0.12 m (input distance converted to meters) Calculate: $$W_i = F_i \times D_i = 50 \times 0.12 = 6 \text{ J}$$ --- 4. **Step b) Calculate Ideal Mechanical Advantage (IMA):** Given: $D_o = 4$ cm = 0.04 m (output distance) Calculate: $$\text{IMA} = \frac{D_i}{D_o} = \frac{0.12}{0.04} = 3$$ --- 5. **Step c) Calculate Efficiency:** First, calculate output work $W_o$: $$W_o = F_o \times D_o = 120 \times 0.04 = 4.8 \text{ J}$$ Calculate Actual Mechanical Advantage (AMA): $$\text{AMA} = \frac{F_o}{F_i} = \frac{120}{50} = 2.4$$ Calculate efficiency: $$\text{Eff} = \frac{W_o}{W_i} = \frac{4.8}{6} = 0.8 = 80\%$$ or equivalently: $$\text{Eff} = \frac{\text{AMA}}{\text{IMA}} = \frac{2.4}{3} = 0.8 = 80\%$$ --- **Final answers:** - a) Input work $W_i = 6$ J - b) Ideal Mechanical Advantage $\text{IMA} = 3$ - c) Efficiency $= 80\%$