Two blocks of masses 10 kg and 20 kg respectively are pushed by a constant appliedforce F across a horizontal frictionless table with constant acceleration such that the blocksremain in contact with e

2.0 m/s².

Explanation:

We can start by finding the net force acting on the system of two blocks. Since there is no friction, the only forces acting on the blocks are the applied force and the force of the 1.0 kg block pushing the 2.0 kg block. The net force can be found using Newton's second law:

F_net = F_applied - F_push

where F_applied is the applied force and F_push is the force of the 1.0 kg block pushing the 2.0 kg block. Plugging in the given values, we get:

F_net = 2.0 N - 2.0 N = 0 N

Since the net force is zero, we know that the acceleration of the system is also zero, according to Newton's second law:

F_net = m_total * a

where m_total is the total mass of the system (1.0 kg + 2.0 kg = 3.0 kg). Solving for a, we get:

a = F_net / m_total = 0 N / 3.0 kg = 0 m/s²

However, we are told that the blocks are actually accelerating with a constant acceleration. This means that there must be another force acting on the system, in addition to the applied force and the force of the 1.0 kg block pushing the 2.0 kg block. This force is the force of the 2.0 kg block pushing back on the 1.0 kg block, according to Newton's third law.

To account for this force, we can use the fact that the two blocks are in contact with each other, which means they must have the same acceleration. Let's call this acceleration a. Then we can write separate equations for each block using Newton's second law:

F_applied - F_push = m_1 * a F_push - F_pull = m_2 * a

where F_pull is the force of the 2.0 kg block pushing back on the 1.0 kg block. We can solve this system of equations for a:

F_applied - F_push - F_pull = (m_1 + m_2) * a 2.0 N - 2.0 N - F_pull = (1.0 kg + 2.0 kg) * a

• F_pull = 3.0 kg * a - 2.0 N F_pull = 2.0 N - 3.0 kg * a

Now we need to find the value of a that satisfies both equations. Substituting the second equation into the first, we get:

F_applied - m_1 * a - (F_pull - m_2 * a) = 0

Plugging in the given values, we get:

2.0 N - 1.0 kg * a - (2.0 N - 2.0 kg * a) = 0

Solving for a, we get:

a = 2.0 m/s²

Therefore, the acceleration of the two blocks is 2.0 m/s².