Question 1:
moderateA block of mass M is placed on a fixed smooth inclined plane of inclination ‘A’ with the horizontal is at rest. The force exerted by the plane on the block has a magnitude:
As the object on inclined plane is at rest net force acting on the object is zero. Force applied on object by earth is Mg. Resultant of Normal reaction and friction force will cancel gravitational force. So the force applied by inclined plane on block is Mg in vertically upward direction.
Question 2:
moderateConsider a car moving on a straight road with a speed of 10 ms-1. The distance at which car can be stopped is: [μ=0.5]Â
Friction force action on the object is f=μ.N= μ.mg
Retardation of the block is a=μ.g = 0.5× 10 =5
Using equation of motion. Stopping distance = u^2/2a=(10 × 10)/( 2* 5)=10 m
Question 3:
moderateA marble block of mass 2 kg lying on ice when given a velocity of 6 ms-1 is stopped by friction in 10 s. Then the coefficient of friction is
Using equation of motion
v= u + at
⇒ 0= 6 + a ×10
⇒ a= -0.6 m/sec^2
Maximum Friction force is μN = μ mg so, a= μ.g
⇒ -0.6 = - μ. 10
⇒μ = 0.06Â
Question 4:
moderateA 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 \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).
Question 5:
moderateA 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 \(s = 0.05\text{ m} = 5\text{ cm}\).
Question 6:
moderateA 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 \( \mu = \frac{2}{3} \).
Question 7:
moderateA conveyor belt is moving at a constant speed of \( 2 \text{ m/s} \). A box is gently dropped on it. The coefficient of friction between them is \( \mu = 0.5 \). The distance that the box will move relative to belt before coming to rest on it, taking \( g = 10 \text{ ms}^{-2} \).
(2011 Mains)
Initial relative speed of box \( u = 2 \text{ m/s} \). Friction force \( f = \mu mg \). Acceleration \( a = \frac{f}{m} = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2 \). Using \( v^2 = u^2 - 2as \) (since friction causes deceleration). With \( v = 0 \), distance \( s = \frac{u^2}{2a} = \frac{(2)^2}{2 \times 5} = \frac{4}{10} = 0.4 \text{ m} \).
Question 8:
moderateA heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is:
(1991)
For equilibrium, the weight of the hanging part must equal the maximum static friction on the table. If (x) is the fraction hanging, then \(xMg = mu(1-x)Mg\). So, \(x = \mu(1-x)\). This gives \(x = \frac{\mu}{1+\mu}\). With \(\mu = 0.25\), \(x = \frac{0.25}{1+0.25} = \frac{0.25}{1.25} = 0.20\) or (20%).