JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given:

• Mass, $m = 5 \, \text{kg}$
• Incline angle, $\theta = 30^\circ$
• Acceleration up the incline, $a = 1 \, \text{m/s}^2$
• Time, $t = 10 \, \text{s}$
• Gravitational acceleration, $g = 10 \, \text{m/s}^2$
• Initial velocity, $u = 0 \, \text{m/s}$

Find: The power delivered by the pulling force at $t = 10 \, \text{s}$.

Along the incline, applying Newton's second law:

$$F - mg \sin\theta = ma$$

Substituting the given values:

$$F - 5 \cdot 10 \cdot 0.5 = 5 \cdot 1$$ $$F - 25 = 5$$ $$F = 30 \, \text{N}$$

Using the first equation of motion:

$$v = u + at$$

Substituting values:

$$v = 0 + 1 \cdot 10 = 10 \, \text{m/s}$$

Power delivered by the pulling force is:

$$P = Fv$$ $$P = 30 \cdot 10 = 300 \, \text{W}$$

Therefore, the power delivered by the pulling force is $300 \, \text{W}$.

Common mistakes

• Using the total weight $mg$ instead of the component $mg\sin\theta$ along the incline is incorrect because only the component parallel to the plane opposes motion. Resolve the weight along the incline before applying Newton's law.

• Assuming power is constant from the start and using an average velocity is wrong here because the question asks for power at $t = 10 \, \text{s}$. First find the instantaneous velocity at that time using $v = u + at$, then use $P = Fv$.

• Forgetting that the plane is smooth leads to introducing friction unnecessarily. Since the incline is smooth, no frictional force acts, so only the pulling force and $mg\sin\theta$ are considered along the plane.