A motor supplies a constant force of 1 kN which is used to move a load a distance of 5 m. The force is then changed to a constant 500 N and the load is moved a further 15 m. Draw the force/distance graph for the operation and from the graph determine the work done by the motor.

Kwanza, tunabadilisha nguvu ya kwanza kutoka kilonewtons (kN) kwenda newtons (N):

$$F_1 = 1 \text{ kN} = 1 \times 1000 \text{ N} = 1000 \text{ N}$$

Umbali wa awamu ya kwanza ni $d_1 = 5 \text{ m}$.
Nguvu ya awamu ya pili ni $F_2 = 500 \text{ N}$.
Umbali wa awamu ya pili ni $d_2 = 15 \text{ m}$.
Jumla ya umbali uliosafirishwa mwishoni mwa awamu ya pili ni $D_{total} = d_1 + d_2 = 5 \text{ m} + 15 \text{ m} = 20 \text{ m}$.

Grafu ya nguvu dhidi ya umbali (force-distance graph) itaonyesha nguvu kwenye mhimili wa y na umbali kwenye mhimili wa x.

• Kuanzia $0 \text{ m}$ hadi $5 \text{ m}$, nguvu ni $1000 \text{ N}$. Hii itawakilishwa na mstari mlalo kutoka $(0, 1000)$ hadi $(5, 1000)$.
• Kuanzia $5 \text{ m}$ hadi $20 \text{ m}$ (yaani, $5 \text{ m} + 15 \text{ m}$), nguvu ni $500 \text{ N}$. Hii itawakilishwa na mstari mlalo kutoka $(5, 500)$ hadi $(20, 500)$.

Kazi iliyofanywa (Work done) ni eneo chini ya grafu ya nguvu dhidi ya umbali. Kwa nguvu isiyobadilika, kazi ni bidhaa ya nguvu na umbali ($W = F \times d$).

Kazi iliyofanywa katika awamu ya kwanza ($W_1$):

$$W_1 = F_1 \times d_1$$ $$W_1 = 1000 \text{ N} \times 5 \text{ m}$$ $$W_1 = 5000 \text{ J}$$

Kazi iliyofanywa katika awamu ya pili ($W_2$):

$$W_2 = F_2 \times d_2$$ $$W_2 = 500 \text{ N} \times 15 \text{ m}$$ $$W_2 = 7500 \text{ J}$$

Jumla ya kazi iliyofanywa na motor ni jumla ya kazi iliyofanywa katika awamu zote mbili:

$$W_{total} = W_1 + W_2$$ $$W_{total} = 5000 \text{ J} + 7500 \text{ J}$$ $$W_{total} = \boxed{12500 \text{ J}}$$