A block of mass 2.10 kg is pushed 2.33 m along a frictionless horizontal table by a constant 19.4 N force directed 27.5o below the horizontal. Determine the magnitude of the normal force exerted by the table.

zero work, no force in direction of motion

To determine the magnitude of the normal force exerted by the table, we need to consider the forces acting on the block.

First, let's determine the vertical component of the applied force. This can be calculated using the trigonometric function sine (sin):

Vertical component = Force * sin(angle)
= 19.4 N * sin(27.5°)

Next, we need to determine the horizontal component of the applied force. This can be calculated using the trigonometric function cosine (cos):

Horizontal component = Force * cos(angle)
= 19.4 N * cos(27.5°)

Since the table is frictionless, the only vertical force acting on the block is the gravitational force, which can be calculated using the mass and acceleration due to gravity:

Gravitational force = Mass * acceleration due to gravity
= 2.10 kg * 9.8 m/s^2

Now, let's analyze the forces acting on the block using Newton's second law of motion:

Sum of forces (in the horizontal direction) = mass * acceleration (in the horizontal direction)

The horizontal forces acting on the block are the applied force and the normal force. Therefore, the equation becomes:

Applied force (horizontal component) + Normal force = mass * acceleration (horizontal)

Substituting the determined values:

19.4 N * cos(27.5°) + Normal force = 2.10 kg * acceleration (horizontal)

Since the block is pushed along the horizontal table, the acceleration in the horizontal direction is 0. Therefore, the equation simplifies to:

19.4 N * cos(27.5°) + Normal force = 0

Now, we can solve for the magnitude of the normal force:

Normal force = -19.4 N * cos(27.5°)

Calculating the value:

Normal force = -19.4 N * cos(27.5°)
= -19.4 N * 0.891
= -17.2894 N

Since the normal force cannot be negative, the magnitude of the normal force exerted by the table is:

Magnitude of normal force = |-17.2894 N|
= 17.2894 N

Therefore, the magnitude of the normal force exerted by the table is approximately 17.2894 N.