# If I have a block that I am pushing up an incline with my hand how do I express the work done on the block by my hand?

You need to know the angle of the incline and the mass of the block to solve this problem, but I'll walk you through the process. I'm making three assumptions:

1. The block in being pushed at a constant rate of speed.

2. The incline is frictionless.

3. The force is being applied to the block in the same direction that it's moving.

Since the block isn't accelerating, the force being exerted on the block to keep it in motion equals the weight of the block acting parallel to the incline. We can find that force vector using trigonometry.

In the diagram below, The arrow pointing down the ramp and parallel to it represents the weight acting in that direction and therefore the magitude of the force being applied. This is what we're trying to calculate. The arrow pointing straight down represents the weight of the block:

w = mg = mass of block x acceleration due to gravity, g = 9.8 m/s^2

Using similar triangles, we can say that the angle between the weight mg and the component of weight perpindicular to the incline is the same as the angle the incline makes with the horizontal. We therefore have a right triangle in which mg is the hypotenuse and the force exerted by your hand is the opposite side.

sin`theta` = opposite side/hypotenuse = F/mg, so F = mg(sin `theta` )

Work = force x displacement where displacement is in the direction of the force.

The work done is therefore expressed as W = mg(sin `theta` )d

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