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------------------------------------------------ A force of 70.0 N is exerted at an angle of 30.0° below the horizontal on a block of mass 8.00 kg that is resting on a table. What is the magnitude of the normal force acting on the block?

A. 43.4 N
B. 78.4 N
C. 113 N
D. 126 N
E. 92.4 N

To determine the magnitude of the normal force acting on the block, we can use the concept of Newton's second law, which states that the sum of the forces in the horizontal direction is equal to the mass of the object times its acceleration in that direction.

Since the block is at rest on the table, the acceleration in the horizontal direction is 0, which means the sum of the forces in the horizontal direction is also 0.

Therefore, the horizontal component of the force applied at an angle of 30.0° below the horizontal is equal in magnitude and opposite in direction to the force of friction.

The vertical component of the force is balanced by the normal force acting on the block.

Since the force of friction is equal in magnitude and opposite in direction to the horizontal component of the applied force, the magnitude of the normal force is equal to the vertical component of the applied force, which can be calculated using the formula:

Normal Force = Applied Force * sin(θ)

Plugging in the values, we have:
Normal Force = 70.0 N * sin(30.0°)
= 35 N

Therefore, the magnitude of the normal force acting on the block is 35 N.

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