A 20kg mass is to be pulled up a slope inclined at 30° to the horizontal. If efficiency of the plane is 75%…

1. Analyze the question

We are given:

• Mass ($m$) = 20 kg
• Angle of inclination ($\theta$) = 30°
• Efficiency ($\eta$) = 75% = 0.75
• Acceleration due to gravity ($g$) = 10 m/s²
• We need to find the Effort ($E$) required to pull the load up the plane.

2. Explain the concept

In an inclined plane, the Work Input is the work done by the effort ($E \times d$), and the Work Output is the work done against gravity to lift the load ($mgh$).

• Efficiency ($\eta$) is defined as: $\text{Efficiency} = \frac{\text{Work Output}}{\text{Work Input}} \times 100\%$
• Work Output = $m \times g \times h$
• Work Input = $E \times L$ (where $L$ is the length of the slope)
• From trigonometry, $\sin \theta = \frac{h}{L}$, which means $h = L \sin \theta$.

Substituting these into the efficiency formula: $\eta = \frac{mgh}{EL} = \frac{mg(L \sin \theta)}{EL} = \frac{mg \sin \theta}{E}$

Therefore, the formula to find the effort ($E$) is: $E = \frac{mg \sin \theta}{\eta}$

3. Show the solution

• Step 1: Calculate the Load ($mg \sin \theta$) Load = $20 \text{ kg} \times 10 \text{ m/s}^2 \times \sin(30^\circ)$ Load = $200 \times 0.5$ Load = $100 \text{ N}$

• Step 2: Calculate the Effort ($E$) using efficiency $E = \frac{\text{Load}}{\text{Efficiency}}$ $E = \frac{100}{0.75}$ $E = \frac{100}{3/4}$ $E = 100 \times \frac{4}{3}$ $E = \frac{400}{3}$ $E = 133.33 \text{ N}$

4. Identify the answer

Comparing our result to the options provided, the correct value is 133.3N. This matches option C.

5. Tip

When dealing with inclined planes, always remember that the "Load" is the component of weight acting down the slope ($mg \sin \theta$). If the machine were 100% efficient, the effort would simply equal the load. Since the efficiency is less than 100%, you must always apply more force than the load, which is why we divide the load by the efficiency (a decimal less than 1).