Irodov Problem 1.122

Suppose that the disc started its journey down the slope from a point s units from the horizontal as shown in the figure. We shall consider the disc's motion on the slope and the horizontal separately.

Motion down the slope: There are three forces acting on the disc along the slope, i) force of gravity mg, ii) the normal reaction and iii) the force of friction opposing its motion . Let the acceleration of the mass down the slope be . Then as shown in the diagram we have,






Motion along the horizontal plane:
Along the horizontal plane force of friction decelerates at a rate w the disc until it stops. The force equations are given below,






The disc started with 0 velocity, achieved maximum velocity at the bend after traveling a distance of s and then stopped after traveling a distance of l. Suppose that the maximum
velocity it achieved was v, then,








The work done by friction along the slope is (the negative sign because the direction of motion is in the opposite direction of the motion). The work done by the force of friction on the horizontal plane is -kmgl. So the total work done by the force of friction is given by,

Irodov Problem 1.121

Consider a certain section of the hill where the slope is an angle . The mass experiences four forces, i) the force of gravity mg acting vertically downwards, ii) the force of friction along the downward slope of the hill since it opposes the upward motion of the mass, iii) the normal reaction from the surface of the hill and iv) the force F directed along the upward slope of the hill.

Since the mass is being dragged "slowly" there the body experiences no acceleration. So basically the force F constantly exactly compensates and cancels out the force of friction Nk and the component of force of gravity along the slope . In other words,



Along the direction normal to the surface there are two forces, i) the normal reaction N and ii) the component of gravity . There is no acceleration along this direction and so we have,



From (1) and (2) we have,



Now the work done dW at some section of the slope by the force when the mass climbs up a distance ds is given by Fds (since the force always aligns itself along the slope). So we have,



From elementary geometry we know that,



From (5a) and (5b) we have,