a block of mass 5 kg rests on a horizontal floor when a constant horizontal force is applied on the block for a time interval 5 second the block slides on a floor India distance 5m under the action of the force and after the removal of force it for the slides a distance 1 metre before coming to stop acceleration due to gravity is 10 M per second magnitude of the applied force is

To determine the magnitude of the applied force on the block, we can break down the problem into two main phases: the sliding phase while the force is applied and the sliding phase after the force is removed. Let's analyze each phase step by step.

Phase 1: Force Applied

During the first phase, a constant horizontal force is applied to the block for 5 seconds, causing it to slide a distance of 5 meters. We can use the equations of motion to find the acceleration of the block.

Using the Equation of Motion

We can use the equation:

• s = ut + (1/2)at²

Where:

• s = distance (5 m)
• u = initial velocity (0 m/s, since it starts from rest)
• a = acceleration (unknown)
• t = time (5 s)

Substituting the known values into the equation:

• 5 = 0 * 5 + (1/2)a(5²)

This simplifies to:

• 5 = (1/2)a(25)

From here, we can solve for acceleration:

• 5 = 12.5a
• a = 5 / 12.5 = 0.4 m/s²

Phase 2: Force Removed

After the force is removed, the block continues to slide for an additional distance of 1 meter before coming to a stop. We need to find the deceleration (negative acceleration) during this phase.

Finding Deceleration

We can use the same equation of motion, but this time we know the final velocity will be 0 m/s when the block stops:

• s = ut + (1/2)at²

Here:

• s = 1 m
• u = final velocity from the first phase (which we need to calculate)
• a = deceleration (unknown)
• t = time (unknown)

First, we need to find the final velocity (u) at the end of the first phase:

• v = u + at

Substituting the known values:

• v = 0 + (0.4)(5) = 2 m/s

Now, we can use this final velocity as the initial velocity for the second phase:

• 1 = (2)t + (1/2)(-a)t²

We need to express time in terms of acceleration. We can use another equation:

• v² = u² + 2as

Setting v = 0, u = 2 m/s, and s = 1 m:

• 0 = (2)² + 2(-a)(1)

This simplifies to:

• 0 = 4 - 2a
• 2a = 4
• a = 2 m/s²

Calculating the Applied Force

Now that we have the acceleration while the force was applied (0.4 m/s²), we can find the net force acting on the block using Newton's second law:

• F_net = ma

Where:

• m = mass of the block (5 kg)
• a = acceleration (0.4 m/s²)

Calculating the net force:

• F_net = 5 kg * 0.4 m/s² = 2 N

However, this is the net force. We also need to account for the force of friction acting against the motion. The force of friction can be calculated using:

• F_friction = μN

Where:

• μ = coefficient of friction (unknown)
• N = normal force (which equals mg = 5 kg * 10 m/s² = 50 N)

Assuming the block is sliding on a surface with a coefficient of friction (let's say μ = 0.2 for this example), we can calculate:

• F_friction = 0.2 * 50 N = 10 N

Now, the total applied force (F_applied) can be found by adding the net force and the force of friction:

• F_applied = F_net + F_friction
• F_applied = 2 N + 10 N = 12 N

Thus, the magnitude of the applied force is 12 N, assuming a coefficient of friction of 0.2. If the coefficient of friction is different, you would need to adjust the calculations accordingly.