Irodov Problem 1.125

The work done by gravity when the body moves a upward vertical distance dy is given by -mgdy (since displacement is opposite to direction of gravity) . The instantaneous power is the derivative of work done and is given by,





a) For the first part the answer is simply 0. This is because before reaching the halfway point the power developed by gravity is always negative (since velocity is upwards and force is downwards). After the halfway point the power generated by gravity is always positive since the body starts to move in the same direction as the force of gravity. For every point A in the left half of the path, there exists a symmetric point A' on the right half of the path where the vertical velocity of body at A is exactly equal and opposite to that at A' so that sum of the instantaneous powers at A and A' will be 0. This means that the negative power accumulated over the first half of the path will be exactly canceled out by the positive power accumulated by the second half of the path.

b) At any instant of time t the vertical velocity of the body is given by (see here for more about projectile motion),

Irodov Problem 1.124

Suppose that some instant x is the length of the chain remaining on the table. The mass of this part of the chain (assuming the chain is homogenous) is given by and hence the force of gravity acting on this piece is .

In the vertical direction there are two forces acting on the chain, i) the force of gravity and ii) the normal reaction from the surface N. There is no acceleration for the chain in the vertical direction and so we have,



Since the chain is slipping, the maximum force of friction that can be offered by the rough surface has been attained and is equal to Nk, here k is the coefficient of friction.

The problem states that only when initially length of the chain hangs, the chain will start slipping. The force of gravity that is responsible for pulling the chain downwards at this initial time is given by . At this time . So from (1) then the force of friction opposing the motion of the chain will be . This means that in order for to be the minimum hanging length to slip we have,






As the chain fall lesser and lesser length of the chain remains on the table. This means that the normal reaction offered by the table surface keeps decreasing as the chain falls. Since the force o friction is given by Nk, even the force of friction keeps decreasing. When x length of the chain remains on the table, the force of friction offered by the surface of the table is given by,




The work done by friction is thus given by,








The negative sign comes because at all points when the chain slips, the force of friction acts in the direction opposed to the motion of the chain.

Irodov Solutions 1.123

Let us suppose the spring constant of the spring is . The tension in the spring when the spring is stretched by a length x is given by .

Let us first consider the mass m2. There are four forces acting on this mass, i) the force of gravity m2g, ii) the normal reaction from the surface N2 iii) the force of friction opposing its motion f2 and iv) the tension in the spring pulling it . Before this mass starts to move there is no acceleration in any direction and so we have,







m2 will break free when f2 attains its maximum possible value i.e. N2k = m2gk. In other words we have for breaking condition,




Now let us consider the mass m1. There are five forces acting on it at any given time, i) the force of gravity m1g, ii) the normal reaction from the surface N1 iii) the force of friction opposing its motion f1 , iv) the tension in the spring pulling it and v) the external force F. The spring can only extend if the friction force f1 has already attained its maximum possible value and can no longer resist the force. In other words,



Considering the forces acting on the mass in the vertical direction and since there is not acceleration in this direction, we have,



In other words,



As soon as the mass m1 breaks free under the force F it will start to accelerate. Suppose at any instant its speed is v as shown in the figure. Then we have,











If you notice (7) is actually nothing but the conservation of energy equation, the total work done by all the forces on the mass m1 gets converted into its kinetic energy. Instead of writing it down, I derived it just to show how it actually originates.

The maximum possible extension of the spring will occur when the mass m1 can no longer move to the left and starts to move back to the right due to the action of the spring. At this point v=0. For the minimum possible value of F this extension will be just sufficient to break m2 free.
From (7) we have,









Last but not the least, thanks to Nalin for showing me the crucial insight into the problem.