Calculating Work Done on a Block Moving Up an Incline

[myk127]

Homework Statement

A block moves up a 30° incline under the action of certain forces, three of which are as follows: F1 is horizontal and magnitude 40N. F2 is normal to the plane and of magnitude 20N. F3 is parallel to the plane and of magnitude 30N. Determine the work done by each force as the block moves 80cm up the incline

Homework Equations

The Attempt at a Solution

I just need to know if I'm doing the right thing here.
WF1 = 40(cos 30)(0.8) = 27.7
WF2 = 20(sin 30)(0.8) = 8
WF3 = 30(0.8) = 24

 

[LowlyPion]

Homework Statement

A block moves up a 30° incline under the action of certain forces, three of which are as follows: F1 is horizontal and magnitude 40N. F2 is normal to the plane and of magnitude 20N. F3 is parallel to the plane and of magnitude 30N. Determine the work done by each force as the block moves 80cm up the incline

Homework Equations

The Attempt at a Solution

I just need to know if I'm doing the right thing here.
WF1 = 40(cos 30)(0.8) = 27.7
WF2 = 20(sin 30)(0.8) = 8
WF3 = 30(0.8) = 24

Welcome to PF.

The first one looks good.

The second one says the force is normal to the plane. If the force is normal to the plane of the incline then there is no work done, because there is no component in the line of motion. I will presume that you are asked about normal to the plane the incline is on. If that is downward then I think you should treat the sin30 component as negative since it is acting against the direction of motion.

 

[myk127]

ah. thanks for that.
what about the F3? you don't have to multiply it to sin 30 becase it's parallel to the inclined right?
so it's like rotating the figure by 30°.

 

[LowlyPion]

ah. thanks for that.
what about the F3? you don't have to multiply it to sin 30 becase it's parallel to the inclined right?
so it's like rotating the figure by 30°.

Correct. It is fully in the direction of the motion. No trig required.