Question

Why is it that using Fs=∆KE to find s gives the same answer as Fs=∆ME for this question:

A mass of 5kg is projected up a smooth slope of 30º with a speed of 6ms^-1. How far up the slope does the mass travel?

Using Fs=∆KE:

Fs=(m/2)*(v^2-u^2)

where m=5, F=-9.8*5*sin30. u=6, v=0

s=3.67m

Using Fs=∆ME:

Fs=∆KE+∆GPE

Fs=(m/2)*(v^2-u^2)+mgh

where m=5, F=-9.8*5*sin30. u=6, v=0, h=s*sin30

Gives the same answer of s=3.67

Why is this?

Answer 1

They don't give the same result the first s is -3.67 while the second is +3.67.
You should have interchanged U and V. The first approach you used is fine for it says You have an initial KE (which must be positive) which will be converted to the work of moving the mass up the inclined plane a distance S.

Answer 2

So to clarify, you're saying that if a body has initial (positive) KE then the work done is equal to the change in KE, regardless of the GPE?

Answer 3

work energy theorem

Net work done on a system = change in K.E of that system

things to rememebr..

its the NET work done.. or work done by NET force.. which means before invoking the W.E theorem.. its your job to calculate how many forces are acting on it.. and find the NET work done..

and yes.. REGARDLESS of any other energy .. works for conservative or non conservative forces.. ALWAYS works!!