a plank of mass 5kg is placed on a frictionless horizontal table. further a block of mass 2kg is placed over the plank. A mass-less spring of natural length 2m is fixed to the plank by its one end. The other end of spring is compressed by the block by half of its springs natural length. The system is now released from the rest. What is the velocity of the plank when block leaves the plank?

To find the velocity of the plank when the block leaves, we need to analyze the forces acting on the system.

First, consider the forces acting on the block. There are two forces acting on it: its weight (mg) and the spring force (kx), where m is the mass of the block, g is the acceleration due to gravity, k is the spring constant, and x is the displacement of the spring from its natural length.

The weight of the block is given by mg, where m = 2 kg and g = 9.8 m/s^2. Therefore, the weight of the block is 2 kg * 9.8 m/s^2 = 19.6 N.

The spring force is given by Hooke's Law: F = kx. In this case, the spring is compressed by half of its natural length, so the displacement x is -1 m (negative because it is compressed). The spring constant k is not given, so we need more information to determine it.

Now let's consider the forces acting on the plank. Since it is placed on a frictionless surface, there is no friction force acting on it. The only force acting on the plank is the spring force from the compressed spring.

Since the spring is attached to the plank, the spring force acting on the block will also act on the plank with an equal magnitude but in the opposite direction. This is according to Newton's third law, which states that for every action, there is an equal and opposite reaction.

Now, to find the velocity of the plank when the block leaves, we need to apply the principle of conservation of momentum. In this case, the momentum of the system is conserved because there are no external forces acting on it. When the block leaves the plank, all the initial momentum goes to the plank.

The momentum of an object is given by p = mv, where p is the momentum, m is the mass of the object, and v is the velocity of the object.

Initially, both the plank and the block are at rest. So the initial momentum is zero for both of them.

When the block leaves, the momentum of the block is also zero since it has a mass of zero at that point. Therefore, all the initial momentum goes to the plank.

Thus, the final momentum of the plank is also zero.

We can use the equation p = mv to solve for the velocity of the plank when the block leaves.

0 = (m_plank + m_block) * v_plank

where m_plank is the mass of the plank and m_block is the mass of the block.

Substituting the given values, we have:

0 = (5 kg + 2 kg) * v_plank

0 = 7 kg * v_plank

Therefore, the velocity of the plank when the block leaves is 0 m/s.

In this case, the compressed spring exerts a force on the block and the plank, but it does not cause any velocity to the plank. The energy stored in the compressed spring is essentially released as potential energy when the block leaves, but as the plank is frictionless, it does not gain any kinetic energy or velocity.