Normal Force on a Brick at Rest

When an object like a brick is placed on a surface, it exerts a downward force on the surface due to its weight. According to Newton's third law, the surface exerts an equal and opposite upward force on the object. This upward force is called the normal force.

In this case, the brick has a mass of 50 kg and is at rest on the floor. Since it is at rest and not moving vertically, the upward normal force is balanced by the downward weight of the brick. The weight ($W$) of the brick can be calculated using the formula:

\(W = m \times g\)

Where:

• \(m\) is the mass of the brick (50 kg)
• \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\))

Calculating the weight:

\(W = 50 \, \text{kg} \times 9.8 \, m/s^2 = 490 \, N\)

Since the brick is at rest, the normal force (\(N\)) exerted by the floor is equal in magnitude to the weight of the brick.

\(N = W = 490 \, N\)

So, the normal force applied by the floor on the brick is 490 N.

Work Done by Normal Force

Work done by a constant force is defined as the dot product of the force and the displacement of the object. Mathematically, it is given by:

\(Work = F \cdot d \cdot \cos(\theta)\)

Where:

• \(F\) is the magnitude of the force
• \(d\) is the magnitude of the displacement
• \(\theta\) is the angle between the force vector and the displacement vector

In this problem, the brick is at rest on the floor without any movement. This means the displacement (\(d\)) of the brick is 0.

The normal force is \(N = 490 \, N\) and it acts vertically upwards. However, since the displacement is zero, the work done by the normal force is:

\(Work = 490 \, N \times 0 \, m \times \cos(\theta)\)

\(Work = 0 \, J\)

Regardless of the direction of the normal force, if the displacement is zero, the work done by the force is always zero.

Therefore, the normal force applied is 490 N and the work done by it is 0 J.