# Work-energy problems on a slope

1. Aug 11, 2010

### jenny egner

1. The problem statement, all variables and given/known data
A box of 2Kg is projected with a speed of 6m/s up a slope at 30* to horizontal. The coeffiecient of friction is 1/3. Use the work energy principle tp calculate the distance travelled by the box before coming to rest.

2. Relevant equations
When you calculate the work done against friction, why don't you take into account the 2gsin30 force acting downwards and only use Fmax? Surely the force acting on the box is both?

3. The attempt at a solution
KE lost = 0.5 x 2 x 36 = 36J
PE gained = 2 x 9.8 x dsin30 = 9.8d

Work done against friction = Fd = (2gsin30 + 1/3 x 2gcos30) x d= 15.46d
Work done = total loss of energy
15.46d = 36 - 9.8d
d=1.43m

2. Aug 11, 2010

### thrill3rnit3

Because the work done against gravity is already taken into account by the change in potential energy.

3. Aug 11, 2010

### Mindscrape

Yeah you actually have it wrong. Work done against friction is
Wf=1/3 * 2gcos30 * d
and work done against gravity is
Wg=2g*d*sin30
which gives you the total work you found. You did it right mathematically, but wrong physically. Doing the physics right is more important than doing the math right.

Also keep in mind that work energy theorem states that change in work equals change in energy, not energy lost equals work done (this is something that is relative).

4. Sep 23, 2010

### setugah

Question.
1. An object is moved in a straight line with a velocity of (3-4t)m/s. A what distance is the object moving in the opposite direction with a velocity of 10m/s?