# Homework Help: Kinetic Energy Problem

1. Dec 15, 2013

### brown20b

1. The problem statement, all variables and given/known data
A 5.0 kg mass is released from rest, at the top of a frictionless 30° inclined plane. Calculate the kinetic energy of the mass when it has slid 40cm down the incline starting from rest. (ANS = 9.80 J)

2. Relevant equations
KE=1/2(mv^2)
a=v/t
d=v1t + 1/2a(t^2)
v2=v1 + at
d= (v1+v2/2)*t

3. The attempt at a solution
G: v1= 0 ; θ= 30° ; d=40cm
R: KE=? ; v2=? ; t=?
(NE direction is +ve)

(1) v2y^2=v1y^2 - 2aydy
v2y^2= √0 -2(9.80)(-0.40sin30)
v2y= 1.98 m/s

(2) v2y=v1y +at
t = v2y/ay
t = 0.202 s

(3) v2x= dx/t
v2x= 0.40cos30/0.202
v2x= 1.71 m/s

(4) v2= √(1.98)^2 + (1.71)^2
v2= 2.62 m/s

(5) KE = 1/2(mv^2)
KE = 1/2(5.0*2.62^2)
KE = 17.2 J

My answer is far off from the correct one so i'm assuming my approach might be wrong.. But I just don't know from where it is that i'm doing it wrong.. Probably something with my calculations, not too sure entirely..

2. Dec 15, 2013

### CAF123

The acceleration down the slope is not 9.8m/s2, what is it?

3. Dec 15, 2013

### brown20b

not sure if it's zero or 9.8sin30

Last edited: Dec 15, 2013
4. Dec 15, 2013

### brown20b

would it be zero?

5. Dec 15, 2013

### azhang40

The acceleration I believe would be Fnet=ma (in the x direction). So add up all the forces in the X direction, which seems to be only one because the surface is frictionless. So mgsin30=5a

6. Dec 15, 2013

### CAF123

brown20b,
Sorry, I misinterpreted your first approach - you have your axes horizontally and vertically rather than perpendicular and parallel to the slope which is the usual orientation of the frame for solving such a problem. But we can still do it your way if you wish.

The axes are usually oriented such that one axis points in the direction of the net acceleration of the block, in this case down the slope and the other axis perpendicular to it. With this orientation, we arrive at the equation posted by azhang40 in one step.

In your attempt, you assumed the acceleration in your y direction was g and the acceleration in your x direction was 0 It is not the case, since the resultant is then g and clearly the block is not in free fall.

I am happy to go through the method with your x and y horizontally and vertically, or would you prefer to consider the reorientation of axes? Both methods follow the same process, one just takes a little longer.