A block of mass 2.50 kg is pushed 2.80 m along a frictionless horizontal table by a constant 10.0 N force directed 25.0° below the horizontal.

1) what's the work done by the applied force? in joules
2)whats the work done by the normal force exerted by the table? in joules

im completely lost on this problem, any help would be appreciated

1) distanceX(ForcepushXcos25)

2) no work - no force in direction of motion
3) no work - no force in direction of motion
4) same as number 1 because table is frictionless and horizontal

To find the work done by a force, you can use the formula:

Work (W) = Force (F) x Distance (d) x Cos θ

where:
- W is the work done by the force,
- F is the magnitude of the force,
- d is the distance over which the force is applied,
- θ is the angle between the force and the direction of motion.

1) Work done by the applied force:

Given:
- The force applied (F) = 10.0 N
- The distance (d) = 2.80 m
- The angle (θ) = 25.0° below the horizontal

First, we need to find the horizontal component of the force. To do this, we can use trigonometry. Since the force is directed 25.0° below the horizontal, the horizontal component can be found using:

Horizontal Component (F_horizontal) = F * Cos θ

Substituting the given values:
F_horizontal = 10.0 N * Cos 25.0°

Now we can calculate the work done by the applied force:

Work = F_horizontal * d

Substituting the values we have:
Work = (10.0 N * Cos 25.0°) * 2.80 m

Now, let's calculate it:

Work = (10.0 N * 0.906) * 2.80 m
Work = 25.13 J (joules)

Therefore, the work done by the applied force is 25.13 joules.

2) Work done by the normal force exerted by the table:

In this case, the normal force exerted by the table is directed perpendicular to the displacement, which means the angle (θ) between the normal force and the direction of motion is 90°.

Since Cos 90° = 0, the work done by the normal force is zero.

Therefore, the work done by the normal force exerted by the table is 0 joules.