Work done walking up a mountain

[ravsterphysics]

Homework Equations

I used F=ma to calculate F and pythag theorem to calculate slope lengths

The Attempt at a Solution

On the left hand side (angle 35) the slope length is 636m, on the right hand side (angle 25) the slope length is 365m.

Now, in my notes I know that F=ma and since there is an angle inbetween, to calculate work done the formula is:

Work done = (F)(Cos,theta)(Distance)

so we should use cos35 and cos 25 but on the videos and mark scheme they use sin instead of cos??

Could someone explain why we use sin and not cos?

 

[Doc Al]

Theta corresponds to what angle? (In the work formula.)

 

[ravsterphysics]

View attachment 110872

Homework Equations

I used F=ma to calculate F and pythag theorem to calculate slope lengths

The Attempt at a Solution

On the left hand side (angle 35) the slope length is 636m, on the right hand side (angle 25) the slope length is 365m.

Now, in my notes I know that F=ma and since there is an angle inbetween, to calculate work done the formula is:

Work done = (F)(Cos,theta)(Distance)

so we should use cos35 and cos 25 but on the videos and mark scheme they use sin instead of cos??

Could someone explain why we use sin and not cos?

Theta corresponds to what angle? (In the work formula.)

the incline angle, so in this case it would be either 35 or 25, that's what I have in my notes

 

[Doc Al]

the incline angle, so in this case it would be either 35 or 25, that's what I have in my notes

In the formula for work, it's the angle between force (what's the force you're working against here?) and the displacement. (Not simply the angle of the incline, though it's surely related.)