Problem 1: Two blocks of 1 kg and 2 kg are connected by a spring as shown in the figure. If spring constant is 500 N/m and coefficient of static and kinetic friction between surfaces are each 0.5, then what minimum constant horizontal force, F, (in Newton) is required to just initiate block on the left ? (Consider g = 10 m / s 2 m / s 2 ).

Figure 1: An external force is applied to initiate block on left.

Block spring system

Solution : The system consists of two blocks and one spring. The system is subjected to an external force. Thus, system is “closed” system, which allows energy exchange with system via work by external force. As friction is also involved, the corresponding energy statement is :

⇒ W E + W F = Δ K + Δ U ⇒ W E + W F = Δ K + Δ U

We are required to find minimum force. We need to understand the implication of this phrase. It is clear that we can apply external force in such a manner that block of “2 kg” acquires kinetic energy by the time block of “1 kg” is initiated in motion. Alternatively as a base case, we can apply external force gradually and slowly in increasing magnitude till the block of “1 kg” is initiated in motion. In this case, the block of “2 kg” does not acquire kinetic energy. This mode of application of external force represents the situation when minimum external force will be required to initiate block of “1 kg”. Hence,

Δ K = 0 Δ K = 0

Motion of block is constrained in horizontal direction. There is no change of vertical elevation. Hence, there is no change in gravitational potential energy. However, spring is stretched from its neutral state. As a consequence, there is change in elastic potential energy of the spring. Let “x” be the extension in the spring by the time block of “1 kg” is initiated in motion. Then, the total change in potential energy is :

Δ U = 1 2 k x 2 Δ U = 1 2 k x 2

The work by friction is done only on the right block of “2 kg” :

W F = - μ m 2 g x W F = - μ m 2 g x

On the other hand, the work by external force is :

W E = F x W E = F x

Putting all these values in the equation of conservation of energy :

⇒ F x − μ m 2 g x = 1 2 k x 2 ⇒ F x − μ m 2 g x = 1 2 k x 2

⇒ F − μ m 2 g = 1 2 k x ⇒ F − μ m 2 g = 1 2 k x

Clearly, we can not solve this equation as there are two unknowns, “F” and “x”. We, therefore, make use of the fact that spring force on the block of mass “1 kg” is equal to maximum static friction,

Figure 2: The spring force is equal to maximum static friction.

Forces on the block

k x = μ m 1 g k x = μ m 1 g

Combining two equations, we have :

⇒ F − μ m 2 g = 1 2 μ m 1 g ⇒ F − μ m 2 g = 1 2 μ m 1 g

⇒ F = μ g m 1 + m 2 2 = 0.5 X 10 1 + 2 2 X 10 = 10 N ⇒ F = μ g m 1 + m 2 2 = 0.5 X 10 1 + 2 2 X 10 = 10 N

Problem 2: In the arrangement shown, the block of mass 10 kg descends through a height of 1 m after being released. The coefficient of friction between block and the horizontal table is 0.3, whereas pulley is friction-less. Considering string and pulley to be “mass-less”, find the speed of the blocks.

Figure 3: Blocks move by 1 m.

Block pulley system

Solution : Here, friction is present as internal force to the system. Hence, we use the form of conservation law as :

W F = Δ K + Δ U W F = Δ K + Δ U

Let us denote 6 kg and 10 kg blocks with subscript “1” and “2”. There is no change of height for the block on the table. Only change in gravitational potential energy is due to change in the height of the block hanging on the other end of the string. Thus, change in potential energy is :

Δ U = m 1 g h + m 2 g h = 0 − 10 x 10 x 1 = - 100 J Δ U = m 1 g h + m 2 g h = 0 − 10 x 10 x 1 = - 100 J

Two blocks are constrained by a taut string. It means that both blocks move with same speed. Let the speed of the blocks after traveling “1 m” be "v". Now, initial kinetic energy of the system is zero. Therefore, the change in kinetic energy is given by :

Δ K = 1 2 m 1 v 2 + 1 2 m 2 v 2 Δ K = 1 2 m 1 v 2 + 1 2 m 2 v 2

⇒ Δ K = 1 2 X 6 X v 2 + 1 2 X 10 X v 2 ⇒ Δ K = 1 2 X 6 X v 2 + 1 2 X 10 X v 2

⇒ Δ K = 8 v 2 ⇒ Δ K = 8 v 2

Friction works only on the block lying on the table. Here, work by friction is given as :

W F = - μ N x = - μ m 1 g x = - 0.3 X 6 X 10 X 1 = - 18 J W F = - μ N x = - μ m 1 g x = - 0.3 X 6 X 10 X 1 = - 18 J

Putting in the equation of conservation of energy,

⇒ - 18 = 8 v 2 − 100 ⇒ - 18 = 8 v 2 − 100

⇒ v 2 = 82 8 ⇒ v 2 = 82 8

⇒ v = 3.2 m / s ⇒ v = 3.2 m / s