Numerical Problems Based on Motion along Rough Inclined Plane for Class 11 Physics

Mass, $m = 2\text{ kg}$; Incline angle, $\theta = 30^\circ$; Coefficient of static friction, $\mu_s = 0.7$. Take $g = 9.8\text{ ms}^{-2}$.

The downward force along the incline pulling the block is:

The maximum static friction (limiting friction) available is:

Note: Since $F_{down} (9.8\text{ N}) < f_{max} (11.9\text{ N})$, the block does not move. In static equilibrium, the actual frictional force only needs to match the applied force to prevent sliding. Therefore, the actual frictional force is $9.8\text{ N}$. However, the textbook answer key ($11.9\text{ N}$) provides the value of the maximum limiting friction.

Mass, $m = 10\text{ kg}$; Angle, $\theta = 30^\circ$; Coefficient of kinetic friction, $\mu_k = 0.5$. Take $g = 9.8\text{ ms}^{-2}$.

The equation of motion for a block sliding down a rough incline is:

Dividing throughout by $m$ gives the acceleration ($a$):

Mass, $m = 10^5\text{ kg}$; Acceleration, $a = 2\text{ ms}^{-2}$; Friction coefficient, $\mu = 0.005$.

An incline of “1 in 50” means $\sin\theta = \frac{1}{50}$. For such small angles, $\cos\theta \approx 1$.

The total force ($F$) required to move the train up must overcome the component of gravity, overcome friction, and provide the net acceleration:

Angle, $\theta = 30^\circ$; Acceleration, $a = \frac{g}{4}$.

The acceleration of a body sliding down a rough inclined plane is given by:

Substitute the given values:

Rearranging to solve for $\mu_k$:

1. Find the Coefficient of Friction ($\mu$):
When the block slides without acceleration (constant velocity), the angle of inclination is the angle of repose. Thus, $\mu = \tan(20^\circ) \approx 0.364$.

2. Calculate New Acceleration ($a$):
When angle is increased to $30^\circ$, the new acceleration is:

3. Calculate Work Done ($W$):
The work done by the net force over distance $s = 1.2\text{ m}$ is:

Mass, $m = 40\text{ metric tons} = 40,000\text{ kg}$; Speed, $v = 54\text{ kmh}^{-1} = 54 \times \frac{5}{18} = 15\text{ ms}^{-1}$.

Incline is “1 in 49”, so $\sin\theta = \frac{1}{49}$. Since the incline is very small, we assume the frictional force ($\mu mg \cos\theta$) is roughly the same as on the level track ($\mu mg$).

Therefore, the additional force required to move up the incline is solely to overcome the downward component of the weight along the slope:

The additional power required is:

Mass, $m = 0.5\text{ kg}$; Angle, $\theta = 30^\circ$; $\mu = 0.2$; $g = 9.8\text{ ms}^{-2}$.

First, calculate the parallel ($mg \sin\theta$) and perpendicular components ($mg \cos\theta$):

• $mg \sin(30^\circ) = 0.5 \times 9.8 \times 0.5 = 2.45\text{ N}$
• Friction, $f = \mu mg \cos(30^\circ) = 0.2 \times 0.5 \times 9.8 \times 0.866 \approx 0.849\text{ N}$

(i) To just prevent sliding down:
Friction acts up the incline, assisting your applied force ($F_1$).

(ii) To just move the block up:
Friction acts down the incline, opposing your applied force ($F_2$).

(iii) To move it up with $a = 20\text{ cms}^{-2} = 0.2\text{ ms}^{-2}$:
Your applied force ($F_3$) must overcome gravity, friction, and provide net acceleration.