Numerical Problems Based on Coefficient of Friction and Angle of Friction for Class 11 Physics

Mass of the block, $m = 1\text{ kg}$; Acceleration of the truck, $a = 5\text{ ms}^{-2}$; Coefficient of static friction, $\mu_s = 0.6$. Let $g = 9.8\text{ ms}^{-2}$.

First, find the maximum static frictional force (limiting friction) available:

Next, find the force required to accelerate the block along with the truck:

Since the required force ($5\text{ N}$) is less than the maximum available static friction ($5.88\text{ N}$), the block does not slip. The actual frictional force simply matches the required force to prevent slipping.

When a body “just slides” down an inclined plane, the angle of inclination ($\theta$) is equal to the angle of repose. At this point, the coefficient of friction ($\mu$) is given by $\mu = \tan\theta$.

The phrase “rises 5 m in every 13 m” refers to the geometry of the incline: Perpendicular (rise) $= 5\text{ m}$, Hypotenuse (distance along incline) $= 13\text{ m}$.

Using the Pythagorean theorem, the base of the triangle is $\sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12\text{ m}$.

Weight of the scooter, $W = 120\text{ kgf}$. The normal reaction ($N$) on a horizontal surface is equal to the weight.

Coefficient of kinetic friction, $\mu = 0.4$. The force of friction ($f$) during skidding is:

Initial velocity, $u = v$; Final velocity, $v_f = 0$ (since it stops).

The maximum retarding force is provided by friction: $F = -\mu N = -\mu mg$.

Using Newton’s second law, the deceleration ($a$) is:

Using the third equation of motion, $v_f^2 – u^2 = 2as$:

This is the required shortest stopping distance.

Initial velocity, $u = 10\text{ ms}^{-1}$; Coefficient of friction, $\mu = 0.4$; Final velocity, $v = 0$. Take $g = 9.8\text{ ms}^{-2}$.

Using the formula derived in the previous question ($s = u^2 / 2\mu g$):

Initial velocity, $u = 7\text{ ms}^{-1}$; Final velocity, $v = 0$; Stopping distance, $s = 10\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

First, find the retardation ($a$) using $v^2 – u^2 = 2as$:

The magnitude of the retarding force (resistance), $F = m|a| = m(2.45)$.

The weight of the car, $W = mg = m(9.8)$.

Now, find the ratio of the resistance to the weight:

Hence proved.

Mass, $m = 3 \times 10^5\text{ kg}$; Constant speed, $v = 50\text{ ms}^{-1}$; Coefficient of friction, $\mu = 0.05$.

To maintain a constant speed, the forward force applied by the engine must exactly balance the frictional resistance ($f$).

Power ($P$) is the product of force and velocity:

Speed, $v = 72\text{ km h}^{-1} = 72 \times \frac{5}{18} = 20\text{ ms}^{-1}$.

Frictional resistance = $50\text{ g wt}$ per quintal.
Total weight = $1000\text{ quintals}$.

Total frictional force ($F$) = $1000 \times 50\text{ g wt} = 50,000\text{ g wt}$.

Convert grams weight to kilograms weight, then to Newtons:

Power ($P$) required to maintain the uniform speed:

Let $m$ be the mass of the car. The normal reaction on the driven rear wheels is $N_{rear} = \frac{1}{2}mg$.

The maximum forward force without slipping is the maximum static friction on the driven wheels:

The maximum acceleration ($a$) of the entire car is given by $f_{max} = ma$:

Target velocity, $v = 100\text{ kmh}^{-1} = 100 \times \frac{5}{18} = 27.78\text{ ms}^{-1}$. Initial velocity $u = 0$.

The phrase “move on the belt before coming to rest” implies finding the relative slipping distance until the suitcase shares the belt’s speed.

Relative to the belt, the suitcase has an initial velocity $u_{rel} = -3\text{ ms}^{-1}$ (moving backwards). It comes to rest relative to the belt, so $v_{rel} = 0$.

The forward friction from the belt provides an acceleration to the suitcase relative to the ground ($a = \mu g = 0.5 \times 9.8 = 4.9\text{ ms}^{-2}$). Relative to the belt, this is a retardation ($a_{rel} = 4.9\text{ ms}^{-2}$ in the positive direction to cancel the negative velocity).

Using the third equation of motion for relative variables ($v_{rel}^2 – u_{rel}^2 = 2a_{rel}s_{rel}$):

(Note: Treating the slip direction as negative, distance magnitude is $s_{rel}$)