A block of mass 2.50 kg is pushed 2.80 m along a frictionless horizontal table by a constant 12.0 N force directed 25.0° below the horizontal.

(a) Determine the work done by the applied force.
J

(b) Determine the work done by the normal force exerted by the table.
J

(c) Determine the work done by the force of gravity.
J

(d) Determine the work done by the net force on the block.
J

To find the work done by a force, you need to multiply the magnitude of the force by the displacement of the object in the direction of the force. In this case, we have multiple forces acting on the block, so we need to consider each one separately.

(a) Work done by the applied force:
The applied force is directed 25.0° below the horizontal, but the displacement is purely horizontal. Since the force is not in the same direction as the displacement, we need to find the component of the force in the direction of displacement. The horizontal component can be found by multiplying the force magnitude by the cosine of the angle:
F_horizontal = F_applied * cos(25.0°)
Work done by the applied force can be calculated as:
Work_applied = F_horizontal * displacement
Substituting the given values:
Work_applied = (12.0 N * cos(25.0°)) * 2.80 m

(b) Work done by the normal force:
The normal force is perpendicular to the displacement, so it does no work. Therefore, the work done by the normal force is zero.

(c) Work done by the force of gravity:
Gravity acts vertically downwards, while the displacement is horizontal. Since there is no vertical displacement, the work done by gravity is also zero.

(d) Work done by the net force on the block:
The net force is the vector sum of the applied force and the force of gravity. Since the force of gravity and the normal force do no work, we only need to consider the work done by the applied force:
Work_net = Work_applied = (12.0 N * cos(25.0°)) * 2.80 m

To find the numerical values of these answers, you can use a calculator to evaluate the trigonometric functions and perform the necessary multiplications.