A block of weight = 25.0 sits on a frictionless inclined plane, which makes an angle = 34.0 with respect to the horizontal, as shown in the figure. A force of magnitude = 14.0 , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

The block moves up an incline with constant speed. What is the total work done on the block by all forces as the block moves a distance = 2.50 up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest.

You have to have units on numbers to give them any meaning.

To find the total work done on the block as it moves up the incline, we need to calculate the work done by the force being applied parallel to the incline.

Work done is equal to the product of force and displacement, where the angle between the force and displacement is the angle between them. In this case, the force and displacement are parallel, so the angle between them is 0 degrees.

Work done = force * displacement * cos(angle)

Given:
Force = 14.0 N
Displacement = 2.50 m
Angle = 0 degrees (force and displacement are parallel)

Plugging in these values into the formula, we have:

Work done = 14.0 N * 2.50 m * cos(0 degrees)

Since cos(0 degrees) = 1, we can simplify the equation:

Work done = 14.0 N * 2.50 m * 1

Therefore, the total work done on the block as it moves up the incline is:

Work done = 35.0 Joules