A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where F is:
The forces acting on the body are: Normal reaction (N(upward) and Friction force f(opposing applied force)
The resultant force is:
A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be \(\ \mu\), then the stopping distance is:
The stopping distance is given by \(s = \frac{v^2}{2a}\), where the frictional retardation is \(a = \mu g\). Substituting the relation for momentum \(v = \frac{P}{m}\) into the formula yields \(s = \frac{P^2}{2\mu m^2g}\).
A particle moves on a rough horizontal ground with some initial velocity say \(v_0\). If \(3/4^{\text{th}}\) of its kinetic energy is lost in friction in time \(t_0\). Then coefficient of friction between the particle and the ground is:
Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).
A body of mass 1 kg has velocity \(1\text{ ms}^{-1}\), up an inclined plane of angle of \(30^\circ\) to the horizontal. The friction coefficient is \(\frac{1}{\sqrt{3}}\). The distance the body travels before stopping is (\(g = 10\text{ m s}^{-2}\)):
Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).
A uniform chain of length \( L \) is placed on a rough horizontal table with some part hanging below the table. If the length of the hanging part becomes \( \frac{2L}{5} \) then the chain starts sliding on the table.
The co-efficient of friction between the chain and the table is:
The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).
A block of mass \(20\text{ kg}\) is placed on a rough horizontal surface, and it is acted upon by a horizontal force of \(40\text{ N}\). If the coefficient of friction is \(0.2\), then the acceleration of the block is
The maximum limiting frictional force is \(f_{max} = \mu m g = 0.2 \times 20 \times 10 = 40\text{ N}\) (using \(g = 10\text{ m/s}^2\)). Since the applied force of \(40\text{ N}\) is equal to the limiting friction, the net horizontal force on the block is zero, resulting in zero acceleration.
Match entries in Column-I with entries in Column-II and choose the correct option.
\(\begin{array}{|l|l|} \hline \text{Column-I} & \text{Column-II} \\[0.5ex] \hline \text{A. Limiting friction} & \text{P. Has magnitude } \mu_k N \\[0.5ex] \text{B. Static friction} & \text{Q. Maximum value of static friction} \\[0.5ex] \text{C. Kinetic friction} & \text{R. Is a self adjusting force} \\[0.5ex] \hline \end{array}\)
Limiting friction is the maximum value of static friction (A \(\rightarrow\) Q). Static friction is a self-adjusting force (B \(\rightarrow\) R). Kinetic friction has a constant magnitude of \(\mu_k N\) (C \(\rightarrow\) P).
Assertion (A): When a person walks on a rough surface, the net force exerted by surface on the person is in the direction of his motion.
Reason (R): Friction force by road on person is against motion.
Assertion (A) is true: To walk, a person pushes backward on the ground. By Newton's third law, the ground exerts a forward static friction force on the person's feet. This forward friction, combined with the normal force, creates a net force from the surface in the direction of motion, propelling the person forward.nReason (R) is false: For walking, the friction force exerted by the road on the person's feet is *in the direction* of the person's motion (forward friction), enabling propulsion. If friction were against the person's motion, walking would be impossible. Therefore, (A) is true but (R) is false.
Assertion (A): The driver of a moving car sees a wall in front of him. To avoid collision, he should apply brakes rather than taking a turn away from the wall.
Reason (R): Friction force is needed to stop the car or taking a turn on a horizontal road.
Assertion (A) is true: In an emergency, applying brakes in a straight line is generally a safer and more controlled maneuver to reduce speed and avoid collision, as sudden turns at high speed can lead to loss of control or skidding.
Reason (R) is true: Both stopping the car (through braking) and taking a turn (requiring centripetal force) on a horizontal road fundamentally rely on the friction force between the tires and the road. Reason (R) is true, but it does not explain *why* braking is preferred over turning; it merely states that friction is involved in both actions.
Therefore, (A) and (R) are true, but (R) is not the correct explanation of (A).
Assertion (A): Walking on horizontal slippery ice can be much more tiring than walking on ordinary pavement.
Reason (R): Walking on ice requires small steps to prevent slipping.
Assertion (A) is true due to low friction on ice. Reason (R) is also true as small steps are needed to minimize horizontal forces and prevent slipping. (R) explains why extra effort and carefulness are required, making the activity tiring.