Find: the force required to move a train of mass \(10^4kg\) up in a inclined plane of 1 in 50 with an acceleration of \(2\frac{m}{s^{2}}\). The co-efficient of friction between train and rails is \(0.005\) and \(g=10\frac{m}{s^{2}}\).
Ans: \(F=\left(2.25\times10^{4}\right)N\).
A box of mass \(5Kg\) is placed on a wooden plank of length \(1.5m\) which is lying on the ground. the plank is lifted from on end along its length so that it becomes inclined. Its being noted that when vertical height of the top end of the plank from the ground becomes \(0.5m\), the box begins to slide. Find: the co-efficient of friction between the box and the plank.
Ans: \(\mu=0.35\).
A block of metal of mass \(50g\) when placed over an inclined plane at an angle of \(15^{o}\) slides down without acceleration. If, the angle of inclination is increased by \(15^{o}\), what will be the acceleration of the metal block over inclined plane?
Ans: \(a=2.6\frac{m}{s^{2}}\).
A mass of \(200kg\) is resting on a rough inclined plane of \(30^{o}\). If, the co-efficient of friction is \(\frac{1}{\sqrt{3}}\), Find: the least and the greatest forces acting parallel to the plane to keep the mass in equilibrium.
Ans:\(F_{min}=Zero\) and \(F_{max}=1960N\).
A coin is placed on an inclined plane which makes an angle \(\theta\) with the horizontal. When, \(\theta\) becomes \(13^{o}\) the coin begins to slide down. What is the co-efficient of static friction between the coin and the inclined plane?
Ans: \(\mu_{s}=0.23\).
A wooden block of mass \(100kg\) rests on flat horizontal wooden floor. The co-efficient of friction between the two being \(0.4\). The block is pulled by a rope making an angle of \(30^{o}\) with the horizontal. What is the minimum tension along the rope that just makes the block sliding?
Ans: \(T_{min}=368N\).
A block ‘A’ of mass \(14kg\) moves along a plane that makes an angle of \(30^{o}\) with the horizontal. Block ‘A’ is connected to another block ‘B’ of mass \(14kg\) by a taut massless string that runs around a frictionless, massless pulley. The block B moves downwards with a constant velocity. Then what is, (i) The magnitude of the frictional force. (ii) The co-efficient of kinetic friction.
Ans:(i) \(f_{k}=68.6N\) and (ii) \(\mu_{k}=0.577]\).
A block slides down an inclined plane of slope \(\theta\) with constant velocity. It is then projected up the same plane with an initial velocity \(v_{o}\). How far up the inclined plane will it move before coming to rest?
Ans: \(S=\frac{v_{o}^{2}}{4gSin\theta}\).
Two blocks are having masses \(4kg\) and \(10kg\) are connected by an ideal string passes over pulley. The block having mass \(4kg\) is free to slide on a surface inclined at an angle \(30^{o}\) with the horizontal whereas \(10kg\) block hangs freely. what will be the acceleration of the system and tension in the string if \(\mu=0.30\).
Ans: \(a=4.872\frac{m}{s^{2}}\) and \(T=49.3N\) .
The upper half of an inclined plane with inclination \(\alpha\) is perfectly smooth, while the lower half is rough. A body starting from the rest at the top will again come to rest at bottom. What will be the co-efficient of friction for the lower half of the inclined plane?
Ans: \(\mu=2tan\alpha\).
The co-efficient of static friction between the block of mass \(2kg\) and table, shown in below figure is \(\mu_{k}=0.2\). What should be the maximum value of \(m\) so that the block do not move? take: \(g=10\frac{m}{s^{2}}\). (string and pulley are light and smooth).
Ans: \(m=0.4kg\).
A car moving with a speed of \(36\frac{km}{hr}\) reaches a upward inclined road of angle of inclination \(30^{0}\), its engine is switched off. What is the maximum distance along the inclined plane moved up by the car before it comes to stop and starts sliding down? Given: Co-efficient of friction is \(0.1\). Acceleration due to gravity = \(10\frac{m}{s^{2}}\).
Ans: \(S=8.524m\).
A body of mass \(M\) is released from the top of a rough inclined plane as shown in below figure. If, the force of friction be \(f\), then prove that the body will reach the bottom with velocity \(v=\sqrt{\frac{2}{M}\left(Mgh-fl\right)}\).
