Work done by gavity on an inclined plane

[fd25t6]

A 30-N box is pulled 6.0 m up along a 37 °inclined plane. What is the work done by the weight (gravitational force) of the box?

Ok so attempt goes like this:

W = FDCOS(37)

So force weight parallel to the plane is 30sin37 = 18.05N
" perppendicular " 30cos37= -23.95


add them up = -5.9n

so -5.9n(6m)cos37 = -28.3 Newtons?

is this right. Thanks in advance for the help

 

[kevinferreira]

This is wrong because you're using the inclination wrong. The work done is defined as
<br /> W=\vec{F}\cdot\Delta\vec{x}<br />
where the force is the weight, 30 Newtons. Try again using some coordinate system (axis along the plane and perpendicular to it, for example), and try again.

More, you don't ever add perpendicular magnitudes of forces, forces are vectors. You can add vectors, but adding magnitudes is meaningless.
Also, the final result is in Joules, or if you prefer 'Newtons meter'.

 

[tiny-tim]

hi fd25t6!

W = FDCOS(37)

correct (except shouldn't it be sin ?) …

work done = force "dot" displacement

the force is the weight, 30N, so why have you done …

So force weight parallel to the plane is 30sin37 = 18.05N
" perppendicular " 30cos37= -23.95

add them up = -5.9n

??

and anyway, you never add the magnitudes of forces in different directions, it's simply not allowed

just apply the equation W = F·D …

what do you get? ​

 

[fd25t6]

Is it really as simple as 30sin37?

 

[tiny-tim]

Is it really as simple as 30sin37?

(minus)

i] work done, like most things in physics, has a formula, and you just apply the formula!

ii] work done = minus potential energy, so how much is the potential energy?

 

[fd25t6]

Well, PE= mgh so 30(6) = -180 J

 

[HallsofIvy]

What, exactly, are you calculating? In the statement of the problem you say "What is the work done" but at the end of your first post you say "-2.83 Newtons". Newtons is a measure of force, not "work" or energy. Work is measured in Joules. A 6 meter ramp at 37 degrees goes up 6sin(37) meters (and over 6 cos(37) meters). The change in potential energy (and so work required to move the box upward) is the force needed, 30g Newtons, times the height: (6 sin(37)(30g)= 180 g sin(37) Joules.

 

[tiny-tim]

no, because the 6 is along the slope, not vertical (and h is vertical)

(and now I'm off to bed :zzz:)