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.
To determine the magnitude of the normal force exerted by the table, we need to consider the forces acting on the block.
1. Weight: The weight of the block can be determined using the formula: weight = mass * acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2.
weight = (2.10 kg) * (9.8 m/s^2) = 20.58 N
2. Horizontal Force: The force applied to push the block horizontally is 19.4 N. However, this force is directed at an angle of 27.5 degrees below the horizontal. We need to determine the horizontal component of this force.
horizontal component of force = applied force * cosine(angle)
horizontal component of force = 19.4 N * cos(27.5 degrees) = 17.49 N
Since there are no other horizontal forces acting on the block (friction is assumed to be negligible), the horizontal force is equal to the horizontal component of the applied force.
3. Normal Force: The normal force is the force exerted by the table perpendicular to its surface. In this case, it counterbalances the weight of the block, making the net vertical force zero. Therefore, the normal force is equal to the weight of the block.
magnitude of normal force = weight = 20.58 N
So, the magnitude of the normal force exerted by the table is 20.58 N.